161 Вправа 161Алгебра 8 клас Тарасенкова 2025 · § 4 · стор. 50

161. Спростіть вираз.

Пам’ятаємо: ${\displaystyle {(b-a)}^2=}$${\displaystyle {(a-b)}^2}$.

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

2) ${\displaystyle \frac{q^3-p^3}{{(p-q)}^2}=}$${\displaystyle \frac{\cancel{\left(q-p\right)}(q^2+qp+p^2)}{\cancel{\left(q-p\right)}(q-p)}=}$${\displaystyle \frac{p^2+pq+q^2}{q-p}}$;

3) ${\displaystyle \frac{2x^2-2y^2}{{(x+y)}^2}=}$${\displaystyle \frac{2(x-y)\cancel{\left(x+y\right)}}{\cancel{\left(x+y\right)}(x+y)}=}$${\displaystyle \frac{2\left(x-y\right)}{x+y}}$;

4) ${\displaystyle \frac{4x^2}{{(2x-y)}^2}-\frac{y^2}{{(2x-y)}^2}=}$${\displaystyle \frac{(2x-y)(2x+y)}{{(2x-y)}^2}=}$${\displaystyle \frac{\cancel{\left(2x-y\right)}(2x+y)}{\cancel{\left(2x-y\right)}(2x-y)}=}$${\displaystyle \frac{2x+y}{2x-y}}$;

5) ${\displaystyle \frac{125-x^3}{{(x-5)}^2}=}$${\displaystyle \frac{\cancel{\left(5-x\right)}(25+5x+x^2)}{\cancel{\left(5-x\right)}(5-x)}=}$${\displaystyle \frac{x^2+5x+25}{5-x}}$;

6) ${\displaystyle \frac{3+x^2+4x+1}{{(x+2)}^3}=}$${\displaystyle \frac{x^2+4x+4}{{(x+2)}^3}=}$${\displaystyle \frac{\cancel{{\left(x+2\right)}^2}}{\cancel{{\left(x+2\right)}^2}(x+2)}=}$${\displaystyle \frac{1}{x+2}}$;

7) ${\displaystyle \frac{32-14x}{{(x-3)}^2}-\frac{26-12x}{{(x-3)}^2}=}$${\displaystyle \frac{32-14x-26+12x}{{(x-3)}^2}=}$${\displaystyle \frac{6-2x}{{(x-3)}^2}=}$${\displaystyle \frac{-2\cancel{\left(x-3\right)}}{\cancel{\left(x-3\right)}(x-3)}=}$${\displaystyle -\frac{2}{x-3}}$;

8) ${\displaystyle \frac{x^2+8y}{(x-4)(y+3)}-\frac{4(x+2y)}{(x-4)(y+3)}=}$${\displaystyle \frac{x^2+8y-4x-8y}{(x-4)(y+3)}=}$${\displaystyle \frac{x\cancel{\left(x-4\right)}}{\cancel{\left(x-4\right)}(y+3)}=}$${\displaystyle \frac{x}{y+3}}$.

(У пункті 8) ${\displaystyle \frac{4(x+2y)}{(4-x)(3+y)}=}$${\displaystyle -\frac{4(x+2y)}{(x-4)(y+3)}}$.)