7 Вправа 7 (повторення)Алгебра 8 клас Тарасенкова 2025 · Повторення · стор. 317

7. Спростіть вираз:

1) ${\displaystyle (\frac{{x-3}^{\mathrm{\setminus }x-3}}{x+3}-\frac{{x+3}^{\mathrm{\setminus }x+3}}{x-3}):\frac{24}{9-x^2}=}$${\displaystyle \frac{{(x-3)}^2-{(x+3)}^2}{(x+3)(x-3)}\cdot \frac{9-x^2}{24}=}$${\displaystyle \frac{\cancel{x^2}-6x\cancel{+9}-\left(\cancel{x^2}+6x\cancel{+9}\right)}{x^2-9}\cdot \frac{-(x^2-9)}{24}=}$${\displaystyle \frac{-12x}{\cancel{x^2-9}}\cdot \frac{\cancel{x^2-9}}{-24}=}$${\displaystyle \frac{\cancel{-12}x}{\underset{2}{\cancel{-24}}}=}$${\displaystyle \frac{x}{2}}$;

2) ${\displaystyle (\frac{{5-x}^{\mathrm{\setminus }x-5}}{x+5}+\frac{{x+5}^{\mathrm{\setminus }x+5}}{x-5}):\frac{20}{25x-x^3}=}$${\displaystyle \frac{-{(x-5)}^2+{(x+5)}^2}{(x+5)(x-5)}\cdot \frac{x(25-x^2)}{20}=}$${\displaystyle \frac{-\left(\cancel{x^2}-10x\cancel{+25}\right)+\cancel{x^2}+10x\cancel{+25}}{x^2-25}\cdot \frac{-x(x^2-25)}{20}=}$${\displaystyle \frac{\stackrel{1}{\cancel{20}}x}{\cancel{x^2-25}}\cdot \frac{x\cancel{\left(x^2-25\right)}}{\underset{1}{\cancel{-20}}}=}$${\displaystyle -x^2}$.