824 Вправа 824Алгебра 8 клас Тарасенкова 2025 · § 20 · стор. 244

824. Спростіть вираз:

1) ${\displaystyle x^2-7x+12=}$${\displaystyle (x-3)(x-4)}$; ${\displaystyle x^2+x-20=}$${\displaystyle (x+5)(x-4)}$;

${\displaystyle (\frac{1}{x^2-7x+12}+\frac{1}{x-3})\cdot \frac{x^2+x-20}{x^2-25}=}$${\displaystyle (\frac{1}{(x-3)(x-4)}+\frac{1}{x-3})\cdot \frac{\cancel{\left(x+5\right)}(x-4)}{(x-5)\cancel{\left(x+5\right)}}=}$${\displaystyle (\frac{1}{(x-3)(x-4)}+\frac{1^{\mathrm{\setminus }x-4}}{x-3})\cdot \frac{x-4}{x-5}=}$${\displaystyle \frac{1+x-4}{(x-3)(x-4)}\cdot \frac{x-4}{x-5}=}$${\displaystyle \frac{\cancel{\left(x-3\right)}}{\cancel{\left(x-3\right)}\cancel{\left(x-4\right)}}\cdot \frac{\cancel{\left(x-4\right)}}{x-5}=}$${\displaystyle \frac{1}{x-5}}$;

2) ${\displaystyle x^2+9x-10=}$${\displaystyle (x+10)(x-1)}$; ${\displaystyle 5x^2-4x-1=}$${\displaystyle (5x+1)(x-1)}$;

${\displaystyle \frac{5x+1}{x^2+9x-10}:\frac{5x^2-4x-1}{x^2-2x+1}+\frac{10}{x^2+10x}=}$${\displaystyle \frac{5x+1}{(x+10)(x-1)}:\frac{(5x+1)(x-1)}{(x-1)(x-1)}+\frac{10}{x(x+10)}=}$${\displaystyle \frac{\cancel{\left(5x+1\right)}}{(x+10)\cancel{\left(x-1\right)}}\cdot \frac{\cancel{\left(x-1\right)}\cancel{\left(x-1\right)}}{\cancel{\left(5x+1\right)}\cancel{\left(x-1\right)}}+\frac{10}{x(x+10)}=}$${\displaystyle \frac{1^{\mathrm{\setminus }x}}{x+10}+\frac{10}{x(x+10)}=}$${\displaystyle \frac{x+10}{x(x+10)}=}$${\displaystyle \frac{\cancel{\left(x+10\right)}}{x\cancel{\left(x+10\right)}}=}$${\displaystyle \frac{1}{x}}$.