228 Вправа 228Алгебра 8 клас Тарасенкова 2025 · § 6 · стор. 66

228. Спростіть вираз.

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

2) ${\displaystyle \frac{10b(b-a)}{9a(a+d)}\cdot \frac{3d\left(d+a\right)}{5ab}=}$${\displaystyle \frac{\cancel{5b}\cdot 2(b-a)\cdot \cancel{3}d\cancel{\left(a+d\right)}}{\cancel{3}\cdot 3a\cancel{\left(a+d\right)}\cdot \cancel{5b}a}=}$${\displaystyle \frac{2d\left(b-a\right)}{3a^2}}$;

3) ${\displaystyle \frac{9{(c-d)}^2}{d^2(c+d)}\cdot \frac{d^4}{3\left(d-c\right)\left(d+c\right)}=}$${\displaystyle \frac{\cancel{3}\cdot 3\cancel{\left(d-c\right)}(d-c)\cancel{d^2}{d}^2}{\cancel{d^2}(c+d)\cdot \cancel{3}\cancel{\left(d-c\right)}(c+d)}=}$${\displaystyle \frac{3d^2\left(d-c\right)}{{\left(c+d\right)}^2}}$;

4) ${\displaystyle {(4-2x)}^3=}$${\displaystyle 8{(2-x)}^3=}$${\displaystyle -8{(x-2)}^3}$; ${\displaystyle {(3-3x)}^2=}$${\displaystyle 9{(x-1)}^2}$;

${\displaystyle \frac{8{(x-2)}^2}{9{(x-1)}^2}\cdot \frac{9{\left(x-1\right)}^2}{-8{\left(x-2\right)}^3}=}$${\displaystyle -\frac{\stackrel{1}{\cancel{8}}\cancel{{\left(x-2\right)}^2}\cdot \cancel{9{\left(x-1\right)}^2}}{\cancel{9{\left(x-1\right)}^2}\cdot \underset{1}{\cancel{8}}\cancel{{\left(x-2\right)}^2}(x-2)}=}$${\displaystyle \frac{1}{2-x}}$.

(У пункті 3) ${\displaystyle {(c-d)}^2=}$${\displaystyle {(d-c)}^2}$.)