202 Вправа 202Алгебра 8 клас Тарасенкова 2025 · § 5 · стор. 59

202. Спростіть вираз.

Спершу розкладаємо чисельники і знаменники на:

1) ${\displaystyle \frac{9c{x}^3}{16ab}\cdot \frac{2a{b}^2}{cxy}\cdot \frac{4b{y}^2}{3a{x}^2}=}$${\displaystyle \frac{9\cdot 2\cdot 4\cdot a{b}^{3}c{x}^3{y}^2}{16\cdot 3\cdot a^{2}bc{x}^{3}y}=}$${\displaystyle \frac{72\cancel{ab}{b}^2\cancel{c{x}^{3}y}y}{48a\cancel{ab}\cancel{c{x}^{3}y}}=}$${\displaystyle \frac{3b^{2}y}{2a}}$;

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

3) ${\displaystyle \frac{a^2-b^2}{a^2+b^2}\cdot \frac{a^4-b^4}{a^2-2ab+b^2}=}$${\displaystyle \frac{(a-b)(a+b)(a^2-b^2)\cancel{\left(a^2+b^2\right)}}{\cancel{\left(a^2+b^2\right)}{(a-b)}^2}=}$${\displaystyle \frac{\cancel{\left(a-b\right)}(a+b)\cancel{\left(a-b\right)}(a+b)}{\cancel{\left(a-b\right)}\cancel{\left(a-b\right)}}=}$${\displaystyle {\left(a+b\right)}^2}$;

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

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

6) ${\displaystyle \frac{x^2+y^2-xy}{a^2-b^2}\cdot \frac{a^2+b^2-2ab}{x^3+y^3}=}$${\displaystyle \frac{\cancel{\left(x^2-xy+y^2\right)}\cancel{\left(a-b\right)}(a-b)}{\cancel{\left(a-b\right)}(a+b)(x+y)\cancel{\left(x^2-xy+y^2\right)}}=}$${\displaystyle \frac{a-b}{\left(a+b\right)\left(x+y\right)}}$.