Transformari identice ale expresiilor algebrice.

Definitie. Vom numi expresie algebrica expresia, ce se obtine din constante si variabile prin intermediul operatiilor de adunare, scadere, inmultire, impartire, ridicare la o putere intreaga si extragerea radacinii.

Exemple de expresii algebrice:

\begin{matrix}E = ^{3} \sqrt{2 + \sqrt{5}}; & E(x, y) = (\frac{x + ^{3} \sqrt{2xy^{2}} − 1}{3\sqrt{z − x} − \sqrt{y}})^{\frac{5}{4}} ; \\ E(x, y, z) = \frac{x + y}{xy} − z; & E(x, y) = (x + y)^{3} − 3xy(x + y). \end{matrix}

Definitie. Domeniu al valorilor admisibile (concis DVA) al expresiei algebrice E(x_{1}, x_{2}, …, x_{n}) (D(E)) se numeste multimea tuturor cortegiilor (x_{1}, x_{2}, …, x_{n}) pentru care expresia E(x_{1}, x_{2}, …, x_{n}) are sens.

De exemplu, DVA al expresiei E(x, y) = \frac{x + y}{xy} − z este multimea D(E) = \{(x, y) | x ∈ R, y ∈ R, xy ≠ 0\}, iar DVA al expresiei E(x, y) = \sqrt{xy} − 2z, este multimea tripletelor {(x, y, z) | x, y, z ∈ R, xy ≥ 0}.

Definitie. Expresiile algebrice E_{1} si E_{2} se numesc identic egale pe multimea M ⊂ D(E_{1}) ∩ D(E_{2}), daca valorile numerice ale acestor expresii sunt egale pentru orice valori ale variabilelor din M.

De exemplu \sqrt{x^{2}} = x pe multimea [0; +∞), \sqrt{a^{2}} = −a, pe multimea (−∞; 0], \frac{x^{2} − 1}{x + 1} = x−1 pe multimea R \ {−1}, (x + y)^{2} = x^{2} + 2xy + y^{2} pe multimea {(x, y) | x ∈ R, y ∈ R}.

Definitie. Vom numi transformare identica a expresiei algebrice pe multimea M ⊆ D(E) inlocuirea acestei expresii cu o expresie identic egala cu ea.

Nota. Tinem sa mentionam, ca uneori multimea M pe care expresiile algebrice sunt identic egale nu se evidentiaza, avandu-se in vedere egalitatea identica a acestor expresii pe intersectia domeniilor lor de definitie.

De exemplu

\sqrt{a^{2}} = |a|, (M = R), \frac{a(a + 1)}{a} = \frac{a^{2} − 1}{a − 1}, (M = R \ \{0; 1\}).

In procesul transformarilor identice sunt utile urmatoarele formule:

0 Copyright\begin{matrix}c \\ ◯ \end{matrix}1999 ONG TCV Scoala Virtuala a Tanarului Matematician http://math.ournet.md

I. Formulele inmultirii prescurtate

1. (a ± b)^{2} = a^{2} ± 2ab + b^{2},

2. (a ± b)^{3} = a^{3} ± 3a^{2}b + 3ab^{2} ± b^{3},

3. a^{2} − b^{2} = (a − b)(a + b),

4. a^{3} ± b^{3} = (a ± b)(a^{2} ∓ ab + b^{2}).

Aceste formule se obtin ca consecinte ale urmatoarelor formule generale:

5. a^{n} − b^{n} = (a − b)(a^{n−1} + a^{n−2}b + … + ab^{n−2} + b^{n−1}) (n ∈ N),

6. a^{2n+1} + b^{2n+1} = (a + b)(a^{2n} − a^{2n−1}b + … − ab^{2n−1} + b^{2n}) (n ∈ N),

7. (binomul lui Newton)

(a + b)^{n} = C_{n}^{0}a^{0}b^{n} + C_{n}^{1}a^{1}b^{n−1} + … + C_{n}^{k}a^{k}b^{n−k} + … + C_{n}^{n}a^{n}b^{0},

unde n ∈ N, C_{n}^{k} = \frac{n!}{k!(n − k)!}, n! = 1 · 2 · 3 · . . . · n, 0! = 1.

II. Proprietatile puterilor

Egalitatile ce urmeaza sunt juste pentru orice numere pozitive a si b si orice numere reale α si β.

1. a^{0} = 1;

2. a^{α+β} = a^{α} · a^{β};

3. a^{α−β} = \frac{a^{α}}{a^{β}};

4. (a^{α})^{β} = a^{αβ};

5. (ab)^{α} = a^{α} · b^{α};

6. (\frac{a}{b})^{α} = \frac{a^{α}}{b^{α}};

0 Copyright\begin{matrix}c \\ ◯ \end{matrix}1999 ONG TCV Scoala Virtuala a Tanarului Matematician http://math.ournet.md

7. a^{−α} = \frac{1}{a^{α}}.

Nota 1. Tinem sa mentionam, ca numerele negative la fel pot fi ridicate la anumite puteri (intregi, sau mai general, rationale de tipul \frac{m}{2n − 1}), unde m - intreg,

Nota 2. 0^{α} = 0, pentru orice α > 0.

III. Proprietatile radicalilor

1. ^{n} \sqrt{a^{n}} = \{ \begin{matrix}a, & daca n − impar, \\ |a|, & daca n − par, \end{matrix}

2. ^{2k} \sqrt{ab} = ^{2k} \sqrt{a} · ^{2k} \sqrt{b}, daca a ≥ 0, b ≥ 0, k ∈ N,

3. ^{2k} \sqrt{ab} = ^{2k} \sqrt{|a|} ^{2k} \sqrt{|b|}, daca ab ≥ 0, k ∈ N.

4. ^{2k+1} \sqrt{ab} = ^{2k+1} \sqrt{a} ^{2k+1} \sqrt{b}, k ∈ N.

5. ( ^{m} \sqrt{a})^{k} = ^{m} \sqrt{a^{k}}, a ≥ 0 daca m este par, a ∈ R daca m este impar.

6. ^{n} \sqrt{\frac{a}{b}} = \begin{matrix}^{n} & \sqrt{a} \\ ^{n} & \sqrt{b} \end{matrix}, daca a ≥ 0, b > 0, n - par sau b ≠ 0, a ∈ R, daca n - impar.

7. ^{n} \sqrt{a^{k}} = a^{\frac{k}{n}} , a ≥ 0.

8. ^{n} \sqrt{\sqrt[m]{a}} = ^{m·n} \sqrt{a}, a ≥ 0 daca m par sau n par, a ∈ R daca m · n - impar.

9. ^{2k+1} \sqrt{−a} = − ^{2k+1} \sqrt{a}, a ∈ R.

10. \sqrt{a ± b\sqrt{c}} = \sqrt{\frac{a + \sqrt{a^{2} − cb^{2}}}{2}} ± \sqrt{\frac{a − \sqrt{a^{2} − cb^{2}}}{2}} , unde a > 0, b > 0, c > 0 si a^{2} ≥ b^{2}c.

Exemplul 1. Sa se determine DVA al expresiilor algebrice:

a) E(x) = ^{6} \sqrt{x + x^{2} − 2x^{3}};

b) E(x, y) = \frac{x^{2} + xy}{x^{2} + y^{2}} (\frac{x}{|x − y|} + \frac{y}{x + y} );

c) E(a, b, c, d) = \frac{a}{b + c} + \frac{\sqrt{d}}{b^{2}c + c^{2}b}.

0 Copyright\begin{matrix}c \\ ◯ \end{matrix}1999 ONG TCV Scoala Virtuala a Tanarului Matematician http://math.ournet.md

Rezolvare. a) DVA al expresiei enuntate se determina rezolvand inecuatia x+x^{2} −2x^{3} ≥ 0 cu ajutorul metodei intervalelor:

x + x^{2} − 2x^{3} ≥ 0 ⇔ x(1 + x − 2x^{2}) ≥ 0 ⇔ x(2x + 1)(1 − x) ≥ 0 ⇔ x ∈ (−∞; −\frac{1}{2}] ∪ [0; 1].

Asadar D(E) = (−∞; −\frac{1}{2}] ∪ [0; 1].

b) Se tine seama ca expresia are sens, daca

\{ \begin{matrix}x^{2} + y^{2} ≠ 0, \\ |x − y| ≠ 0, \\ x + y ≠ 0, \end{matrix}

de unde rezulta D(E) = {(x, y) | x ≠ y, x ≠ −y}.

c) Cum numitorii fractiilor rationale urmeaza a fi diferiti de zero, iar radacina de ordinul doi exista numai din expresii nenegative, se obtine sistemul

\{ \begin{matrix}b + c ≠ 0, \\ b^{2}c + c^{2}b ≠ 0, \\ d ≥ 0, \end{matrix} ⇔ \{ \begin{matrix}b + c ≠ 0, \\ bc(b + c) ≠ 0, \\ d ≥ 0, \end{matrix} ⇔ \{ \begin{matrix}b + c ≠ 0, \\ b ≠ 0, \\ c ≠ 0, \\ d ≥ 0. \end{matrix}

Asadar DVA al expresiei date este multimea {(a, b, c, d) | b + c ≠ 0, b ≠ 0, c ≠ 0, d ≥ 0}.

Exemplul 2. Sa se determine daca expresiile A si B sunt identic egale pe multimea M.

\begin{matrix}a) A = (\frac{a\sqrt{a} + b\sqrt{b}}{\sqrt{a} + \sqrt{b}} − \sqrt{ab})^{\frac{1}{2}} \frac{\sqrt{a} + \sqrt{b}}{ab(a − b)}, B = \frac{1}{ab}, M = \{(a, b) | a > b > 0\}; \\ b)A = \frac{a − \sqrt{3}}{\sqrt{(\frac{a^{2}+3}{2a})^{2} − 3}}, B = \frac{2a}{a + \sqrt{3}}, M = \{a | a > \sqrt{3}\}. \end{matrix}

Rezolvare. a) Cum a\sqrt{a} = (\sqrt{a})^{3}, b\sqrt{b} = (\sqrt{b})^{3}, a = (\sqrt{a})^{2}, b = (\sqrt{b})^{2} pe mulitmea M, se aplica formulele inmultirii prescurtate si se obtine:

0 Copyright\begin{matrix}c \\ ◯ \end{matrix}1999 ONG TCV Scoala Virtuala a Tanarului Matematician http://math.ournet.md

A = (\frac{(\sqrt{a})^{3} + (\sqrt{b})^{3}}{\sqrt{a} + \sqrt{b}} − \sqrt{ab})^{\frac{1}{2}} · \frac{\sqrt{a} + \sqrt{b}}{ab((\sqrt{a})^{2} − (\sqrt{b})^{2})} =

= (\frac{(\sqrt{a} + \sqrt{b})((\sqrt{a})^{2} − \sqrt{ab} + (\sqrt{b})^{2})}{\sqrt{a} + \sqrt{b}} − \sqrt{ab})^{\frac{1}{2}} · \frac{(\sqrt{a} + \sqrt{b})}{ab(\sqrt{a} − \sqrt{b})(\sqrt{a} + \sqrt{b})} =

= [(\sqrt{a})^{2} − 2\sqrt{ab} + (\sqrt{b})^{2}]^{\frac{1}{2}} · \frac{1}{ab(\sqrt{a} − \sqrt{b})} = \sqrt{(\sqrt{a} − \sqrt{b})^{2}} \frac{1}{ab(\sqrt{a} − \sqrt{b})} =

= |\sqrt{a} − \sqrt{b}| · \frac{1}{ab(\sqrt{a} − \sqrt{b})},

deoarece a > b > 0 implica \sqrt{a} > \sqrt{b}, si prin urmare |\sqrt{a} − \sqrt{b}| = \sqrt{a} − \sqrt{b}. Asadar A = (\sqrt{a} − \sqrt{b})\frac{1}{ab(\sqrt{a} − \sqrt{b})} = \frac{1}{ab} = B. Astfel pe multimea M expresiile A si B sunt egale.

b) Similar exemplului precedent

A = \frac{a − \sqrt{3}}{\sqrt{(\frac{a^{2} + 3}{2a})^{2} − 3}} = \frac{a − \sqrt{3}}{\sqrt{\frac{(a^{2} + 3)^{2} − 4a^{2} · 3}{4a^{2}}}} = \frac{(a − \sqrt{3}) · 2|a|}{\sqrt{(a^{2} − 3)^{2}}} = \frac{(a − \sqrt{3}) · 2|a|}{|a^{2} − 3|} = \frac{2a}{a + \sqrt{3}} = B.

S-a tinut seama ca daca a > \sqrt{3}, atunci \sqrt{a^{2}} = a si \sqrt{(a^{2} − 3)^{2}} = |a^{2} − 3| = a^{2} − 3.

0 Copyright\begin{matrix}c \\ ◯ \end{matrix}1999 ONG TCV Scoala Virtuala a Tanarului Matematician http://math.ournet.md

Exemplul 3. Sa se aduca la o forma mai simpla expresiile:

a) \frac{\sqrt{2b + 2\sqrt{b^{2} − 4}}}{\sqrt{b^{2} − 4} + b + 2};

b) ( \frac{^{3} \sqrt{mn^{2}} + ^{3} \sqrt{m^{2}n}}{^{3} \sqrt{m^{2}} + 2 ^{3} \sqrt{mn} + ^{3} \sqrt{n^{2}}} − 2 ^{3} \sqrt{n} + \frac{m − n}{^{3} \sqrt{m^{2}} − ^{3} \sqrt{n^{2}}} ) \frac{1}{^{6} \sqrt{m} + ^{6} \sqrt{n}};

c) \frac{( \frac{^{4} \sqrt{bc^{3}} + ^{4} \sqrt{a^{2}bc}}{\sqrt{a} + \sqrt{c}} + ^{4} \sqrt{bc})^{2} + bc + 3}{\sqrt{bc} + 3};

d) \frac{a^{3}}{(a − b)(a − c)} + \frac{b^{3}}{(b − c)(b − a)} + \frac{c^{3}}{(c − a)(c − b)};

e) \frac{a^{2}b^{2}}{(a − c)(b − c)} + \frac{a^{2}c^{2}}{(a − b)(c − b)} + \frac{b^{2}c^{2}}{(b − a)(c − a)};

f) \frac{y − z}{(x − y)(x − z)} + \frac{z − x}{(y − x)(y − z)} + \frac{x − y}{(z − x)(z − y)};

g) \frac{m|m − 3|}{(m^{2} − m − 6)|m|}.

Rezolvare. DVA al expresiei se determina din sistemul \{ \begin{matrix}b^{2} − 4 ≥ 0, \\ 2b + 2\sqrt{b^{2} − 4} ≥ 0, \\ \sqrt{b^{2} − 4} + b + 2 > 0, \end{matrix} de unde

rezulta b ≥ 2.

In DVA expresia este identic egala cu:

\begin{matrix}\frac{\sqrt{2b + 2\sqrt{b^{2} − 4}}}{\sqrt{b^{2} − 4} + b + 2} = \frac{\sqrt{(\sqrt{b − 2})^{2} + (\sqrt{b + 2})^{2} + 2\sqrt{b − 2}\sqrt{b + 2}}}{\sqrt{b − 2}\sqrt{b + 2} + (\sqrt{b + 2})^{2}} = \\ = \frac{\sqrt{(\sqrt{b − 2} + \sqrt{b + 2})^{2}}}{\sqrt{b + 2}(\sqrt{b − 2} + \sqrt{b + 2})} = \frac{|\sqrt{b − 2} + \sqrt{b + 2}|}{\sqrt{b + 2}(\sqrt{b − 2} + \sqrt{b + 2})} = \frac{1}{\sqrt{b + 2}}, \end{matrix}

deoarece in DVA \sqrt{b − 2} + \sqrt{b + 2} ≥ 2 si prin urmare |\sqrt{b − 2} + \sqrt{b + 2}| = \sqrt{b − 2} + \sqrt{b + 2}. Asadar, pentru b ≥ 2 expresia enuntata este egala cu \frac{1}{\sqrt{b + 2}}.

b) DVA al expresiei este multimea \{(m, n) | m ≥ 0, n ≥ 0, m ≠ n\}. Se noteaza ^{6} \sqrt{m} = a, 6 \sqrt{n} = b, atunci ^{3} \sqrt{m} = a^{2}, ^{3} \sqrt{m^{2}} = a^{4}, m = a^{6} si ^{3} \sqrt{n} = b^{2}, ^{3} \sqrt{n^{2}} = b^{4}, n = b^{6} si expresia devine

0 Copyright\begin{matrix}c \\ ◯ \end{matrix}1999 ONG TCV Scoala Virtuala a Tanarului Matematician http://math.ournet.md

(\frac{a^{2}b^{4} + a^{4}b^{2}}{a^{4} + 2a^{2}b^{2} + b^{4}} − 2b^{2} + \frac{a^{6} − b^{6}}{a^{4} − b^{4}} ) \frac{1}{a + b} =

= (\frac{a^{2}b^{2}(a^{2} + b^{2})}{(a^{2} + b^{2})^{2}} − 2b^{2} + \frac{(a^{2} − b^{2})(a^{4} + a^{2}b^{2} + b^{4})}{(a^{2} − b^{2})(a^{2} + b^{2})}) \frac{1}{(a + b)} =

= \frac{a^{2}b^{2} − 2b^{2}(a^{2} + b^{2}) + a^{4} + a^{2}b^{2} + b^{4}}{a^{2} + b^{2}} · \frac{1}{a + b} =

= \frac{a^{4} − b^{4}}{a^{2} + b^{2}} · \frac{1}{a + b} = \frac{(a^{2} + b^{2})(a − b)(a + b)}{(a^{2} + b^{2})(a + b)} = a − b.

Asadar expresia initiala pe DVA este identic egala cu ^{6} \sqrt{m} − ^{6} \sqrt{n}. c) In DVA : {(a, b, c) | a ≥ 0, b ≥ 0, c ≥ 0, a^{2} + c^{2} ≠ 0} expresia se tranforma astfel:

\frac{( \frac{^{4} \sqrt{bc^{3}} + ^{4} \sqrt{a^{2}bc}}{\sqrt{a} + \sqrt{c}} + ^{4} \sqrt{bc})^{2} + bc + 3}{\sqrt{bc} + 3} = \frac{( \frac{^{4} \sqrt{bc}(\sqrt{c} + \sqrt{a})}{\sqrt{a} + \sqrt{c}} + ^{4} \sqrt{bc})^{2} + bc + 3}{\sqrt{bc} + 3} =

= \frac{4\sqrt{bc} + bc + 3}{\sqrt{bc} + 3} = \frac{3\sqrt{bc} + 3 + \sqrt{bc} + bc}{\sqrt{bc} + 3} = \frac{3(\sqrt{bc} + 1) + (1 + \sqrt{bc})\sqrt{bc}}{\sqrt{bc} + 3} =

= \frac{(\sqrt{bc} + 1)(\sqrt{bc} + 3)}{\sqrt{bc} + 3} = \sqrt{bc} + 1.

d) DVA al expresiei este multimea {(a, b, c) | a ≠ b, a ≠ c, b ≠ c}. Se aduce expresia la numitor comun si se obtine

\frac{a^{3}}{(a − b)(a − c)} + \frac{b^{3}}{(b − c)(b − a)} + \frac{c^{3}}{(c − a)(c − b)} = \frac{a^{3}(c − b) + b^{3}(a − c) + c^{3}(b − a)}{(a − b)(a − c)(c − b)}.

Se tine seama de expresia de la numitor si se descompune in factori numaratorul a^{3}(c − b) + b^{3}(a − c) + c^{3}(b − a) = c(a^{3} − b^{3}) + ab(b^{2} − a^{2}) + c^{3}(b − a) = = (a − b)(c(a^{2} + ab + b^{2}) − ab(a + b) − c^{3}) = (a − b)(c(a^{2} − c^{2}) + ab(c − a) + b^{2}(c − a)) = = (b − c)(a − b)(−a^{2}b − a^{2}c + c^{2}(a + b)) = (a − b)(b − c)(b(c^{2} − a^{2}) + ac(c − a)) = = (a − b)(b − c)(c − a)(ab + bc + ca). Asadar pe DVA expresia enuntata este identic egala cu ab + bc + ca.

0 Copyright\begin{matrix}c \\ ◯ \end{matrix}1999 ONG TCV Scoala Virtuala a Tanarului Matematician http://math.ournet.md

f) DVA al expresiei este multimea {(x, y, z) | x ≠ y, y ≠ z, z ≠ x}. Primul termen al expresiei se descompune in modul urmator:

\frac{y − z}{(x − y)(x − z)} = \frac{x − z − x + y}{(x − y)(x − z)} = \frac{x − z}{(x − y)(x − z)} − \frac{x − y}{(x − y)(x − z)} = \frac{1}{x − y} − \frac{1}{x − z} .

Similar se descompun si ceilalti termeni

\begin{matrix}\frac{z − x}{(y − z)(y − x)} = \frac{1}{y − z} − \frac{1}{y − x}; \\ \frac{x − y}{(z − x)(z − y)} = \frac{1}{z − x} − \frac{1}{z − y} . \end{matrix}

Prin urmare

\frac{y − z}{(x − y)(x − z)} + \frac{z − x}{(y − z)(y − x)} + \frac{x − y}{(z − x)(z − y)} =

= \frac{1}{x − y} − \frac{1}{x − z} + \frac{1}{y − z} − \frac{1}{y − x} + \frac{1}{z − x} − \frac{1}{z − y} =

= \frac{2}{x − y} + \frac{2}{y − z} + \frac{2}{z − x}.

g) DVA al expresiei este R \ {−2; 0; 3}. Se tine seama ca expresia contine |m| si |m − 3| si se considera urmatoarele trei cazuri:

1. fie m ∈ (−∞; −2) ∪ (−2; 0). Atunci |m| = −m, |m − 3| = −(m − 3) si expresia devine

\frac{m|m − 3|}{(m^{2} − m − 6)|m|} = \frac{−m(m − 3)}{−(m + 2)(m − 3)m} = \frac{1}{m + 2};

2. fie m ∈ (0; 3). Atunci |m| = m, |m − 3| = −(m − 3) si expresia devine

\frac{m|m − 3|}{(m^{2} − m − 6)|m|} = \frac{−m(m − 3)}{(m + 2)(m − 3)m} = −\frac{1}{m + 2};

3. fie m ∈ (3; +∞). Atunci |m| = m, |m − 3| = m − 3 si expresia devine

\frac{m|m − 3|}{(m^{2} − m − 6)|m|} = \frac{1}{m + 2}.

Asadar

\frac{m|m − 3|}{(m^{2} − m − 6)|m|} = \{ \begin{matrix}\frac{1}{m + 2}, & daca m ∈ (−∞; −2) ∪ (−2; 0) ∪ (3; +∞), \\ −\frac{1}{m + 2}, & daca m ∈ (0; 3). \end{matrix}

0 Copyright\begin{matrix}c \\ ◯ \end{matrix}1999 ONG TCV Scoala Virtuala a Tanarului Matematician http://math.ournet.md

Exemplul 4. Sa se descompuna in factori

a) (x + y)(y + z)(z + x) − xyz;
b) x^{3} + y^{3} + z^{3} − 3xyz;
c) x^{8} + x^{7} + x^{6} + x^{5} + x^{4} + x^{3} + x^{2} + x + 1;
d) x^{5} + x + 1.

Rezolvare. a) Se aduna si se scade z(y + z)(z + x) si se grupeaza convenabil

(x + y)(y + z)(z + x) + z(y + z)(z + x) − z(y + z)(z + x) − xyz =
= (y + z)(z + x)(x + y + z) − z((y + z)(z + x) − xy) =
= (y + z)(z + x)(x + y + z) − z(z^{2} + yz + zx) =
= (y + z)(z + x)(x + y + z) − z^{2}(x + y + z) =
= (x + y + z)((y + z)(z + x) − z^{2}) = (x + y + z)(xy + yz + zx).

b) Se aplica formula pentru suma a doua cuburi si se rezolva similar exemplului precedent:

x^{3} + y^{3} + z^{3} − 3xyz = (x + y)(x^{2} − xy + y^{2}) + z(z^{2} − 3xy) =
= (x + y + z)(x^{2} − xy + y^{2}) + z(z^{2} − 3xy − x^{2} + xy − y^{2}) =
= (x + y + z)(x^{2} − xy + y^{2}) + z(z^{2} − (x + y)^{2}) =
= (x + y + z)(x^{2} − xy + y^{2} + z(z − x − y)) =
= (x + y + z)(x^{2} + y^{2} + z^{2} − xy − xz − yz).

c) Se aplica formulele inmultirii prescurtate si se obtine

\begin{matrix}x^{8} + x^{7} + x^{6} + x^{5} + x^{4} + x^{3} + x^{2} + x + 1 = \frac{x^{9} − 1}{x − 1} = \frac{(x^{3})^{3} − 1^{3}}{x − 1} = \\ = \frac{(x^{3} − 1)(x^{6} + x^{3} + 1)}{x − 1} = (x^{2} + x + 1)(x^{6} + x^{3} + 1). \end{matrix}

d) x^{5}+x+1 = 1+x+x^{2}−x^{2}+x^{5} = 1+x+x^{2}−x^{2}(1−x^{3}) = (1+x+x^{2})−x^{2}(1−x)(1+x+x^{2}) = = (1 + x + x^{2})(1 − x^{2}(1 − x)) = (1 + x + x^{2})(1 − x^{2} + x^{3}).

0 Copyright\begin{matrix}c \\ ◯ \end{matrix}1999 ONG TCV Scoala Virtuala a Tanarului Matematician http://math.ournet.md

Exemplul 5. Sa se rationalizeze numitorii urmatoarelor expresii irationale:

a) \frac{1}{1 + \sqrt{2} − \sqrt{3}}; b) \frac{1}{\sqrt{3} + \sqrt{5} + \sqrt{7}}; c) \frac{1}{1 + ^{3} \sqrt{2} + 2 ^{3} \sqrt{4}}.

Rezolvare. Se multiplica cu conjugata numitorului si se obtine:

a) \frac{1}{1 + \sqrt{2} − \sqrt{3}} = \frac{1 + \sqrt{2} + \sqrt{3}}{(1 + \sqrt{2})^{2} − 3} = \frac{1 + \sqrt{2} + \sqrt{3}}{2\sqrt{2}} = \frac{(1 + \sqrt{2} + \sqrt{3})\sqrt{2}}{2\sqrt{2} · \sqrt{2}} = \frac{\sqrt{2} + 2 + \sqrt{6}}{4}.

b) Similar exemplului a) se obtine

\frac{1}{\sqrt{3} + \sqrt{5} + \sqrt{7}} = \frac{\sqrt{3} + \sqrt{5} − \sqrt{7}}{(\sqrt{3} + \sqrt{5})^{2} − 7} = \frac{\sqrt{3} + \sqrt{5} − \sqrt{7}}{1 + 2\sqrt{15}} =

= \frac{(\sqrt{3} + \sqrt{5} − \sqrt{7})(2\sqrt{15} − 1)}{(2\sqrt{15})^{2} − 1} = \frac{(\sqrt{3} + \sqrt{5} − \sqrt{7})(2\sqrt{15} − 1)}{59}.

c) Se utilizeaza formula (a se vedea exemplul 4 b)):

x^{3} + y^{3} + z^{3} − 3xyz = (x + y + z)(x^{2} + y^{2} + z^{2} − xy − yz − zx)

de unde

x + y + z = \frac{(x^{3} + y^{3} + z^{3} − 3xyz)}{(x^{2} + y^{2} + z^{2} − xy − yz − zx)}

si se obtine

\frac{1}{1 + ^{3} \sqrt{2} + 2 ^{3} \sqrt{4}} = \frac{1^{2} + ( ^{3} \sqrt{2})^{2} + (2 ^{3} \sqrt{4})^{2} − ^{3} \sqrt{2} − 2 ^{3} \sqrt{4} − 2 ^{3} \sqrt{8}}{1^{3} + ( ^{3} \sqrt{2})^{3} + (2 ^{3} \sqrt{4})^{3} − 3 · 1 · ^{3} \sqrt{2} · 2 · ^{3} \sqrt{4}} = \frac{7 ^{4} \sqrt{2} − ^{3} \sqrt{4} − 3}{23};

Exemplul 6. Sa se arate ca numarul

a) \sqrt{|20\sqrt{7} − 53|} − \sqrt{20\sqrt{7} + 53},
b) ^{3} \sqrt{26 − 15\sqrt{3}} + \sqrt{3},
c) \sqrt{26 + 6\sqrt{13 − 4\sqrt{8 + 2\sqrt{6 − 2\sqrt{5}}}}} + \sqrt{26 − 6\sqrt{13 + 4\sqrt{8 − 2\sqrt{6 + 2\sqrt{5}}}}},

este numar intreg. Rezolvare. a) Se evidentiaza sub radicali patrate complete si se obtine:

\sqrt{(5 − 2\sqrt{7})^{2}} − \sqrt{(5 + 2\sqrt{7})^{2}} = |5 − 2\sqrt{7}| − |5 + 2\sqrt{7}| = 2\sqrt{7} − 5 − 5 − 2\sqrt{7} = −10.

0 Copyright\begin{matrix}c \\ ◯ \end{matrix}1999 ONG TCV Scoala Virtuala a Tanarului Matematician http://math.ournet.md

b) Se evidentiaza sub semnul radacinii de ordinul trei un cub complet si se obtine 3 \sqrt{26 − 15\sqrt{3}} + \sqrt{3} = ^{3} \sqrt{8 − 12\sqrt{3} + 18 − 3\sqrt{3}} + \sqrt{3} = ^{3} \sqrt{(2 − \sqrt{3})^{3}} + \sqrt{3} = 2 − \sqrt{3} + \sqrt{3} = 2.

c) Se tine seama, ca 6 ± 2\sqrt{5} = 5 ± 2\sqrt{5} + 1 = (\sqrt{5} ± 1)^{2}, \sqrt{8 + 2\sqrt{6 − 2\sqrt{5}}} = \sqrt{8 + 2\sqrt{5} − 2} = \sqrt{6 + 2\sqrt{5}} = \sqrt{5} + 1, \sqrt{8 − 2\sqrt{6 + 2\sqrt{5}}} = \sqrt{8 − 2\sqrt{5} − 2} = \sqrt{6 − 2\sqrt{5}} = \sqrt{5} − 1, \sqrt{13 − 4(\sqrt{5} + 1)} = \sqrt{9 − 4\sqrt{5}} = \sqrt{5 − 2 · 2 · \sqrt{5} + 4} = \sqrt{5} − 2, \sqrt{13 + 4(\sqrt{5} − 1)} = \sqrt{5} + 2, \sqrt{26 + 6(\sqrt{5} − 2)} = \sqrt{14 + 6\sqrt{5}} = \sqrt{5 + 2 · 3 · \sqrt{5} + 9} = \sqrt{5} + 3, \sqrt{26 − 6\sqrt{\sqrt{5} + 2}} = \sqrt{14 − 6\sqrt{5}} = \sqrt{(\sqrt{5} − 3)^{2}} = |\sqrt{5} − 3| = 3 − \sqrt{5}

si se obtine ca expresia initiala este egala cu \sqrt{5} + 3 + 3 − \sqrt{5} = 6.

Identitati conditionate.

Exemplul 7. a) Sa se calculeze x^{2} + y^{2} + z^{2}, daca x + y + z = 1, \frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 0.

b) Sa se arate ca egalitatea xyz = 1 implica

\frac{1}{1 + x + xy} + \frac{1}{1 + y + yz} + \frac{1}{1 + z + zx} = 1.

c) Sa se arate ca daca x + y + z = 0, atunci x^{4} + y^{4} + z^{4} = 2(xy + yz + zx)^{2}.

d) Sa se demonstreze, ca pentru orice trei termeni consecutivi ai unei progresii geometrice are loc egalitatea

a_{1}^{2} + a_{2}^{2} + a_{3}^{2} = (a_{1} + a_{2} + a_{3})(a_{1} − a_{2} + a_{3}).

0 Copyright\begin{matrix}c \\ ◯ \end{matrix}1999 ONG TCV Scoala Virtuala a Tanarului Matematician http://math.ournet.md

e) Sa se arate ca daca x_{1} + x_{2} + x_{3} = y_{1} + y_{2} + y_{3} = x_{1}y_{1} + x_{2}y_{2} + x_{3}y_{3} = 0 si nu toate numerele x_{j}, j = \overline{1, 3} si y_{i}, i = \overline{1, 3} sunt egale cu zero, atunci

\frac{x_{1}^{2}}{x_{1}^{2} + x_{2}^{2} + x_{3}^{2}} + \frac{y_{1}^{2}}{y_{1}^{2} + y_{2}^{2} + y_{3}^{2}} = \frac{2}{3}.

Rezolvare. a) Cum egalitatea \frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 0 implica xy + yz + zx = 0, rezulta

x^{2} + y^{2} + z^{2} = (x + y + z)^{2} − 2(xy + yz + zx) = 1^{2} − 2 · 0 = 1.

b) Se observa, ca in conditia xyz = 1

\frac{1}{1 + z + zx} = \frac{1}{xyz + z + zx} = \frac{1}{z(1 + x + xy)} = \frac{xy}{1 + x + xy} ,

si atunci

\frac{1}{1 + y + yz} = \frac{1}{y(1 + z + zx)} = \frac{1}{y · z(1 + x + xy)} = \frac{x}{1 + x + xy} .

Prin urmare,

\frac{1}{1 + x + xy} + \frac{1}{1 + y + yz} + \frac{1}{1 + z + zx} = \frac{1 + x + xy}{1 + x + xy} = 1.

c) Se tine seama ca x + y = −z si se obtine

2(xy + yz + zx)^{2} = 2(xy + z(x + y))^{2} = 2(xy − (x + y)^{2})^{2} =
= 2(x^{2} + xy + y^{2})^{2} = 2(x^{4} + x^{2}y^{2} + y^{4} + 2x^{3}y + 2x^{2}y^{2} + 2xy^{3}) =
= x^{4} + y^{4} + (y^{4} + 4xy^{3} + 6x^{2}y^{2} + 4yx^{3} + x^{4}) = x^{4} + y^{4} + (x + y)^{4} =
= x^{4} + y^{4} + (−z)^{4} = x^{4} + y^{4} + z^{4}.

d) Cum a_{1}a_{3} = a_{2}^{2} (proprietatea caracteristica a progresiei geometrice) rezulta

(a_{1} + a_{2} + a_{3})^{2} = a_{1}^{2} + a_{2}^{2} + a_{3}^{2} + 2a_{2}(a_{1} + a_{2} + a_{3}).

Prin urmare

a_{1}^{2} + a_{2}^{2} + a_{3}^{2} = (a_{1} + a_{2} + a_{3})^{2} − 2a_{2}(a_{1} + a_{2} + a_{3}) = (a_{1} + a_{2} + a_{3})(a_{1} − a_{2} + a_{3}).

0 Copyright\begin{matrix}c \\ ◯ \end{matrix}1999 ONG TCV Scoala Virtuala a Tanarului Matematician http://math.ournet.md

e) Se considera vectorii (x_{1}, x_{2}, x_{3}), (y_{1}, y_{2}, y_{3}) si (1, 1, 1). Cum produsurile scalare ale acestor vectori sunt egale cu zero, rezulta ca ei sunt doi cate doi ortogonali. Asadar vectorii

g_{1} = (\frac{x_{1}}{\sqrt{x_{1}^{2} + x_{2}^{2} + x_{3}^{2}}} , \frac{x_{2}}{\sqrt{x_{1}^{2} + x_{2}^{2} + x_{3}^{2}}} , \frac{x_{3}}{\sqrt{x_{1}^{2} + x_{2}^{2} + x_{3}^{2}}} ) ,

g_{2} = (\frac{y_{1}}{\sqrt{y_{1}^{2} + y_{2}^{2} + y_{3}^{2}}} , \frac{y_{2}}{\sqrt{y_{1}^{2} + y_{2}^{2} + y_{3}^{2}}} , \frac{y_{3}}{\sqrt{y_{1}^{2} + y_{2}^{2} + y_{3}^{2}}} ) ,

g_{3} = (\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}) ,

formeaza o baza ortonormata in spatiul R^{3}. Prin urmare

(x, g_{1})^{2} + (x, g_{2})^{2} + (x, g_{3})^{2} = 1,

pentru orice vector x ∈ R^{3}.

In particular, pentru x = e_{1} = (1, 0, 0), se obtine

\frac{x_{1}^{2}}{x_{1}^{2} + x_{2}^{2} + x_{3}^{2}} + \frac{y_{1}^{2}}{y_{1}^{2} + y_{2}^{2} + y_{3}^{2}} + \frac{1}{3} = 1,

de unde rezulta egalitatea initiala.

Exercitii pentru autoevaluare

1. Sa se determine DVA al expresiei

\begin{matrix}a) y = \sqrt{3x − x^{2}} & R: [0; 3]. \\ b) y = \sqrt{\frac{x^{2} − 7x + 12}{x^{2} − 2x − 3}} & R: (−∞; −1) ∪ [4; +∞). \\ c) y = \frac{(b + c)(a + c)}{ab + cd + cb + ad} & R: \{(a, b, c) | a ≠ −c, b ≠ −d\}. \end{matrix}

2. Sa se determine, daca expresiile A si B sunt identic egale pe multimea M, daca:

A = \frac{a − \sqrt{3}}{\sqrt{(\frac{a^{2} + 3}{2a}^{2} − 3)}}, B = \frac{2a}{a + \sqrt{3}}, M = \{a : a > \sqrt{3}\}.

0 Copyright\begin{matrix}c \\ ◯ \end{matrix}1999 ONG TCV Scoala Virtuala a Tanarului Matematician http://math.ournet.md

R: Da.

3. Sa se aduca la forma mai simpla expresia

a) (\frac{\sqrt{a + 1}}{2} − \frac{1}{2\sqrt{a + 1}}) · (\frac{\sqrt{a + 1} − 1}{\sqrt{a + 1} + 1} − \frac{\sqrt{a + 1} + 1}{\sqrt{a + 1} − 1}) · \frac{\sqrt{a + 1}}{a};  R: − 1
b) (a + \frac{ab}{a − b}) (\frac{ab}{a + b} − a) : \frac{a^{2} + b^{2}}{a^{2} − b^{2}} ;  R: − \frac{a^{4}}{a^{2} + b^{2}}
c) \frac{\sqrt{x} + 1}{1 + \sqrt{x} + x} : \frac{1}{x^{2} − \sqrt{x}};  R: \sqrt{x}(x − 1)
d) \frac{ab}{a^{\frac{2}{3}} − a^{\frac{1}{3}} b^{\frac{1}{3}} + b^{\frac{2}{3}}} − \frac{a − b}{a^{\frac{2}{3}} + a^{\frac{1}{3}} b^{\frac{1}{3}} + b^{\frac{2}{3}}} − \frac{a^{\frac{2}{3}} − b^{\frac{2}{3}}}{a^{\frac{1}{3}} − b^{\frac{1}{3}}} ;  R: b^{\frac{1}{3}} − a^{\frac{1}{3}}

e) ^{4} \sqrt{(1 − 2a + a^{2})(a^{2} − 1)(a − 1)} : \frac{a^{2} + 2a − 3}{\sqrt[4]{a + 1}};

R: \frac{\sqrt{a + 1}}{a + 3} pentru a > 1; −\frac{\sqrt{a + 1}}{a + 3} pentru − 1 < a < 1.

f) \frac{a^{2} − 4 − |a − 2|}{a^{3} + 2a^{2} − 5a − 6};

R: \frac{1}{a + 3} pentru a > 2; \frac{1}{a + 2} pentru a < 2, a ≠ −3, a ≠ −1.

g) \frac{b − c}{b + c} + \frac{c − a}{c + a} + \frac{a − b}{a + b} + \frac{(b − c)(c − a)(a − b)}{(b + c)(c + a)(a + b)};  R: 0.

4. Sa se descompuna in factori:

\begin{matrix}a) x^{4} + x^{2} + 1; & R: (x^{2} + x + 1)(x^{2} − x + 1) \\ b) (x − y)^{3} + (y − z)^{3} + (z − x)^{3}; & R: 3(x − y)(y − z)(z − x). \end{matrix}

0 Copyright\begin{matrix}c \\ ◯ \end{matrix}1999 ONG TCV Scoala Virtuala a Tanarului Matematician http://math.ournet.md

5. Sa se rationalizeze

\begin{matrix}a) \frac{9}{\sqrt{10} + \sqrt{15} + \sqrt{14} + \sqrt{21}}; & R: \frac{9(\sqrt{7} − \sqrt{5})(\sqrt{3} − \sqrt{2})}{2}. \\ b) \frac{\sqrt{7}}{^{12} \sqrt{5} + ^{12} \sqrt{3}}; & R: \frac{\sqrt{7}( ^{12} \sqrt{5} − ^{12} \sqrt{3})( ^{6} \sqrt{5} + ^{6} \sqrt{3})( ^{3} \sqrt{25} + ^{3} \sqrt{15} + ^{3} \sqrt{9})}{2}. \end{matrix}

6. Sa se arate ca expresiile

\begin{matrix}a) \frac{\sqrt{2 ^{4} \sqrt{8} − 2\sqrt{\sqrt{2} + 1}}}{\sqrt{^{4} \sqrt{8} + \sqrt{\sqrt{2} − 1}} − \sqrt{^{4} \sqrt{8} − \sqrt{\sqrt{2} − 1}}}; & R: 1 \\ b) ^{3} \sqrt{9 + \sqrt{80}} + ^{3} \sqrt{9 − \sqrt{80}}; & R: 3 \end{matrix}

reprezinta numere intregi.

0 Copyright\begin{matrix}c \\ ◯ \end{matrix}1999 ONG TCV Scoala Virtuala a Tanarului Matematician http://math.ournet.md