4 Вправа 4 (повторення)Алгебра 8 клас Тарасенкова 2025 · Повторення · стор. 316

4. Скоротіть дріб:

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

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

3) ${\displaystyle x^2-3x+2=}$${\displaystyle 0}$: ${\displaystyle x_1+x_2=}$${\displaystyle 3}$; ${\displaystyle x_1\cdot x_2=}$${\displaystyle 2}$; ${\displaystyle x_1=}$${\displaystyle 1}$; ${\displaystyle x_2=}$${\displaystyle 2}$; ${\displaystyle x^2-3x+2=}$${\displaystyle (x-1)(x-2)}$;

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

4) ${\displaystyle x^2-3x-4=}$${\displaystyle 0}$: ${\displaystyle D=}$${\displaystyle {(-3)}^2-4\cdot 1\cdot (-4)=}$${\displaystyle 9+16=}$${\displaystyle 25}$; ${\displaystyle x_1=}$${\displaystyle \frac{3-5}{2}=}$${\displaystyle -1}$; ${\displaystyle x_2=}$${\displaystyle \frac{3+5}{2}=}$${\displaystyle 4}$; ${\displaystyle x^2-3x-4=}$${\displaystyle (x+1)(x-4)}$;

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