393° Вправа 393Алгебра 8 клас Тарасенкова 2025 · § 10 · стор. 109

393. Спростіть вираз:

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

2) ${\displaystyle \frac{2}{a}+a^{-1}=}$${\displaystyle \frac{2}{a}+\frac{1}{a}=}$${\displaystyle \frac{3}{a}}$;

3) ${\displaystyle \frac{3}{m^2}+m^{-2}=}$${\displaystyle \frac{3}{m^2}+\frac{1}{m^2}=}$${\displaystyle \frac{4}{m^2}}$;

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

5) ${\displaystyle {(a+2)}^{-1}-\frac{2}{a+2}=}$${\displaystyle \frac{1}{a+2}-\frac{2}{a+2}=}$${\displaystyle \frac{1-2}{a+2}=}$${\displaystyle -\frac{1}{a+2}}$;

6) ${\displaystyle a{(a+b)}^{-1}+b{(a+b)}^{-1}=}$${\displaystyle \frac{a}{a+b}+\frac{b}{a+b}=}$${\displaystyle \frac{\cancel{a+b}}{\cancel{a+b}}=}$${\displaystyle 1}$;

7) ${\displaystyle {(1+b)}^0-{(1+b)}^{-1}=}$${\displaystyle 1-\frac{1}{1+b}=}$${\displaystyle \frac{1^{\mathrm{\setminus }1+b}}{1}-\frac{1}{1+b}=}$${\displaystyle \frac{\stackrel{1}{\cancel{1}}+b\cancel{-1}}{1+b}=}$${\displaystyle \frac{b}{1+b}}$;

8) ${\displaystyle {(a+b)}^{-1}+{(a+b)}^0=}$${\displaystyle \frac{1}{a+b}+1=}$${\displaystyle \frac{1}{a+b}+\frac{1^{\mathrm{\setminus }a+b}}{1}=}$${\displaystyle \frac{1+a+b}{a+b}}$;

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

10) ${\displaystyle {(5+b)}^{-1}+{(5-b)}^{-1}=}$${\displaystyle \frac{1^{\mathrm{\setminus }5-b}}{5+b}+\frac{1^{\mathrm{\setminus }5+b}}{5-b}=}$${\displaystyle \frac{1\cdot \left(5-b\right)}{(5+b)(5-b)}+\frac{1\cdot \left(5+b\right)}{(5+b)(5-b)}=}$${\displaystyle \frac{\left(5-b\right)+\left(5+b\right)}{(5+b)(5-b)}=}$${\displaystyle \frac{5\cancel{-b}+5\cancel{+b}}{25-b^2}=}$${\displaystyle \frac{10}{25-b^2}}$.