287 Вправа 287Алгебра 8 клас Бевз 2025 · § 6 · стор. 60

287. Заповніть таблицю.

Спростимо вирази:

${\displaystyle \frac{x^2-2x+1}{x+1}:\frac{x-1}{{(x+1)}^2}=}$${\displaystyle \frac{x^2-2x+1}{x+1}\cdot \frac{{(x+1)}^2}{x-1}}$=

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

=${\displaystyle (x-1)(x+1)=}$${\displaystyle x^2-1}$;

допустимі ${\displaystyle x\ne -1}$, ${\displaystyle x\ne 1}$;

${\displaystyle \frac{3x^2-48}{x-1}:\frac{12+3x}{1-x}=}$${\displaystyle \frac{3x^2-48}{x-1}\cdot \frac{1-x}{12+3x}}$=

=${\displaystyle \frac{(3x^2-48)\cdot (1-x)}{(x-1)\cdot (12+3x)}=}$${\displaystyle -\frac{\cancel{3}\cancel{\left(x+4\right)}\cancel{\left(x-1\right)}(x-4)}{\cancel{3}\cancel{\left(x+4\right)}\cancel{\left(x-1\right)}}}$=

=${\displaystyle -(x-4)=}$${\displaystyle 4-x}$;

допустимі ${\displaystyle x\ne 1}$, ${\displaystyle x\ne -4}$.