159.

Zadatak

TEKST ZADATKA

Srediti izraz:

(aa+b+1a−b+a+b−b2b2−a2)⋅a2+2ab+b2a3+b3(\frac{a}{a+b}+\frac{1}{a-b}+\frac{a+b-b^2}{b^2-a^2}) \cdot \frac{a^2+2ab+b^2}{a^3+b^3}
REŠENJE ZADATKA

Primeniti formulu za kvadrat binoma: (a+b)2=a2+2ab+b2(a+b)^2=a^2+2ab+b^2 i formulu za zbir kubova: a3+b3=(a+b)(a2−ab+b2)a^3+b^3=(a+b)(a^2-ab+b^2)

(aa+b+1a−b+a+b−b2b2−a2)⋅(a+b)2(a+b)(a2−ab+b2)(\frac{a}{a+b}+\frac{1}{a-b}+\frac{a+b-b^2}{b^2-a^2}) \cdot \frac{{(a+b)^2}}{{(a+b)}(a^2-ab+b^2)}

Skraiti zajednički činilac: (a+b).(a+b).

(aa+b+1a−b+a+b−b2b2−a2)⋅a+ba2−ab+b2(\frac{a}{a+b}+\frac{1}{a-b}+\frac{a+b-b^2}{b^2-a^2}) \cdot \frac{{a+b}}{a^2-ab+b^2}

Izvući minus ispred zagrade:

(aa+b+1a−b − ◯a+b−b2a2−b2)⋅a+ba2−ab+b2(\frac{a}{a+b}+\frac{1}{a-b} \textcircled{\ -\ } \frac{a+b-b^2}{a^2-b^2}) \cdot \frac{{a+b}}{a^2-ab+b^2}

Primeniti formulu za razliku kvadrata: (a2−b2)=(a−b)(a+b)(a^2-b^2)=(a-b)(a+b)

(aa+b+1a−b−a+b−b2(a−b)(a+b))⋅a+ba2−ab+b2(\frac{a}{a+b}+\frac{1}{a-b} - \frac{a+b-b^2}{(a-b)(a+b)}) \cdot \frac{{a+b}}{a^2-ab+b^2}

Izraz u zagradi svesti na jedan razlomak.

a(a−b)+(a+b)−(a+b−b2)(a−b)(a+b)⋅a+ba2−ab+b2\frac{a(a-b)+(a+b)-(a+b-b^2)}{(a-b)(a+b)}\cdot \frac{{a+b}}{a^2-ab+b^2}

Osloboditi se zagrada.

a2−ab+a+b−a−b+b2(a−b)(a+b)⋅a+ba2−ab+b2\frac{a^2-ab+a+b-a-b+b^2}{(a-b)(a+b)}\cdot \frac{{a+b}}{a^2-ab+b^2}

Srediti izraz.

a2−ab+b2(a−b)(a+b)⋅a+ba2−ab+b2\frac{a^2-ab+b^2}{(a-b)(a+b)}\cdot \frac{{a+b}}{a^2-ab+b^2}
a2−ab+b2(a−b)(a+b)⋅a+ba2−ab+b2\frac{\cancel{a^2-ab+b^2}}{(a-b)\cancel{(a+b)}}\cdot \frac{{\cancel{a+b}}}{\cancel{a^2-ab+b^2}}
1a−b\frac{1}{a-b}

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