165 Вправа 165Алгебра 8 клас Тарасенкова 2025 · § 4 · стор. 50

165. Виконайте дії.

1) ${\displaystyle \frac{(4y-2)\cdot 35+(3+3y)\cdot 21+(1-3y)\cdot 15}{525}=}$${\displaystyle \frac{140y-70+63+63y+15-45y}{525}=}$${\displaystyle \frac{158y+8}{525}}$;

2) ${\displaystyle xy-x^2=}$${\displaystyle x(y-x)}$; ${\displaystyle xy-y^2=}$${\displaystyle -y(y-x)}$;

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

3) (як у № 164, 4) ${\displaystyle \frac{1}{t\left(t-1\right)}}$;

4) ${\displaystyle t^2-9=}$${\displaystyle (t-3)(t+3)}$; ${\displaystyle t^2+3t=}$${\displaystyle t(t+3)}$;

${\displaystyle \frac{4\cdot t-2\cdot \left(t-3\right)}{t(t-3)(t+3)}=}$${\displaystyle \frac{4t-2t+6}{t(t-3)(t+3)}=}$${\displaystyle \frac{2\cancel{\left(t+3\right)}}{t(t-3)\cancel{\left(t+3\right)}}=}$${\displaystyle \frac{2}{t\left(t-3\right)}}$.