311 Вправа 311Алгебра 8 клас Бевз 2025 · § 7 · стор. 65

311. Спрости вираз.

а) ${\displaystyle (\frac{a^2}{b^2}+\frac{b}{a}):(\frac{a}{b^2}-\frac{1}{b}+\frac{1}{a})}$;

1) ${\displaystyle \frac{{a^2}^{\mathrm{\setminus }a}}{b^2}+\frac{b^{\mathrm{\setminus }{b}^2}}{a}=}$${\displaystyle \frac{a^3+b^3}{a{b}^2}=}$${\displaystyle \frac{(a+b)(a^2-ab+b^2)}{a{b}^2}}$;

2) ${\displaystyle \frac{a^{\mathrm{\setminus }a}}{b^2}-\frac{1^{\mathrm{\setminus }ab}}{b}+\frac{1^{\mathrm{\setminus }{b}^2}}{a}=}$${\displaystyle \frac{a^2-ab+b^2}{a{b}^2}}$;

3) ${\displaystyle \frac{(a+b)(a^2-ab+b^2)}{a{b}^2}:\frac{a^2-ab+b^2}{a{b}^2}}$=

=${\displaystyle \frac{(a+b)(a^2-ab+b^2)}{a{b}^2}\cdot \frac{a{b}^2}{a^2-ab+b^2}}$=

=${\displaystyle \frac{\cancel{a}\cancel{b^2}(a+b)\cancel{\left(a^2-ab+b^2\right)}}{\cancel{a}\cancel{b^2}\cancel{\left(a^2-ab+b^2\right)}}=}$${\displaystyle a+b}$;

б) ${\displaystyle \frac{3a^4}{a^2+x^2-2ax}:{(1+\frac{x}{a-x})}^2}$;

1) ${\displaystyle 1+\frac{x}{a-x}=}$${\displaystyle \frac{1^{\mathrm{\setminus }x-a}}{1}-\frac{x}{x-a}=}$${\displaystyle \frac{x-a-x}{x-a}}$=

=${\displaystyle \frac{-a}{x-a}=}$${\displaystyle \frac{a}{a-x}}$;

2) ${\displaystyle {\left(\frac{a}{a-x}\right)}^2=}$${\displaystyle \frac{a^2}{{(a-x)}^2}}$;

3) ${\displaystyle \frac{3a^4}{a^2+x^2-2ax}:\frac{a^2}{{(a-x)}^2}=}$${\displaystyle \frac{3a^4}{a^2+x^2-2ax}\cdot \frac{{(a-x)}^2}{a^2}}$=

=${\displaystyle \frac{3a^4\cdot {(a-x)}^2}{(a^2+x^2-2ax)\cdot a^2}=}$${\displaystyle \frac{3\cancel{a^2}{a}^2\cancel{{\left(a-x\right)}^2}}{\cancel{a^2}\cancel{{\left(a-x\right)}^2}}=}$${\displaystyle 3a^2}$.