417* Вправа 417Алгебра 8 клас Тарасенкова 2025 · § 10 · стор. 113

417. Спростіть вираз:

1) ${\displaystyle \frac{{(1+x)}^{-1}}{x+2}+\frac{{(2+x)}^{-1}}{x+3}+\frac{{(3+x)}^{-1}}{x+4}=}$${\displaystyle \frac{1}{(x+1)(x+2)}+\frac{1}{(x+2)(x+3)}+\frac{1}{(x+3)(x+4)}}$; ${\displaystyle \frac{1}{(x+1)(x+2)}=}$${\displaystyle \frac{1}{x+1}-\frac{1}{x+2}}$ (так само інші):

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

2) ${\displaystyle (\frac{a^{-1}+b^{-1}}{a^{-1}-b^{-1}}-\frac{a^{-1}-b^{-1}}{a^{-1}+b^{-1}}):(\frac{a^2+b^2}{a^2-b^2}-\frac{a^2-b^2}{a^2+b^2})}$;

перша дужка (як у № 409.4): ${\displaystyle \frac{4ab}{b^2-a^2}}$; друга дужка:

${\displaystyle \frac{{(a^2+b^2)}^2-{(a^2-b^2)}^2}{(a^2-b^2)(a^2+b^2)}=}$${\displaystyle \frac{4a^2{b}^2}{a^4-b^4}}$;

${\displaystyle \frac{4ab}{b^2-a^2}:\frac{4a^2{b}^2}{a^4-b^4}=}$${\displaystyle \frac{\cancel{4ab}\cdot (a^2+b^2)\cancel{\left(a^2-b^2\right)}}{-\cancel{\left(a^2-b^2\right)}\cdot \cancel{4ab}\cdot ab}=}$${\displaystyle -\frac{a^2+b^2}{ab}}$;

3) ${\displaystyle \frac{{(a{b}^{-1}+1)}^2}{a{b}^{-1}-a^{-1}b}\cdot \frac{a^3{b}^{-3}-1}{a^2{b}^{-2}+a{b}^{-1}+1}:\frac{a^2{b}^{-2}+1}{a{b}^{-1}+a^{-1}b}}$;

нехай ${\displaystyle t=}$${\displaystyle a{b}^{-1}=}$${\displaystyle \frac{a}{b}}$, тоді ${\displaystyle a^{-1}b=}$${\displaystyle \frac{1}{t}}$:

${\displaystyle \frac{{(t+1)}^2}{t-\frac{1}{t}}\cdot \frac{t^3-1}{t^2+t+1}:\frac{t^2+1}{t+\frac{1}{t}}=}$${\displaystyle \frac{t{(t+1)}^2}{(t-1)(t+1)}\cdot (t-1):\frac{t(t^2+1)}{t^2+1}=}$${\displaystyle \frac{\cancel{t}(t+1)\cancel{\left(t-1\right)}}{\cancel{\left(t-1\right)}\cdot \cancel{t}}=}$${\displaystyle t+1=}$${\displaystyle \frac{a+b}{b}}$.