394° Вправа 394Алгебра 8 клас Тарасенкова 2025 · § 10 · стор. 110

394. Спростіть вираз:

1) ${\displaystyle \frac{1}{y^2}-y^{-2}=}$${\displaystyle \frac{1}{y^2}-\frac{1}{y^2}=}$${\displaystyle 0}$;

2) ${\displaystyle {(a+10)}^{-1}+\frac{9}{a+10}=}$${\displaystyle \frac{1}{a+10}+\frac{9}{a+10}=}$${\displaystyle \frac{10}{a+10}}$;

3) ${\displaystyle a{(a+4)}^{-1}+4{(a+4)}^{-1}=}$${\displaystyle \frac{a}{a+4}+\frac{4}{a+4}=}$${\displaystyle \frac{\cancel{a+4}}{\cancel{a+4}}=}$${\displaystyle 1}$;

4) ${\displaystyle \frac{7}{7+n}+n{(7+n)}^{-1}=}$${\displaystyle \frac{7}{7+n}+\frac{n}{7+n}=}$${\displaystyle \frac{\cancel{7+n}}{\cancel{7+n}}=}$${\displaystyle 1}$;

5) ${\displaystyle {(m+n)}^{-1}-{(m+n)}^0=}$${\displaystyle \frac{1}{m+n}-1=}$${\displaystyle \frac{1}{m+n}-\frac{1^{\mathrm{\setminus }m+n}}{1}=}$${\displaystyle \frac{1-m-n}{m+n}}$;

6) ${\displaystyle {(4+x)}^{-1}-{(4-x)}^{-1}=}$${\displaystyle \frac{1^{\mathrm{\setminus }4-x}}{4+x}-\frac{1^{\mathrm{\setminus }4+x}}{4-x}=}$${\displaystyle \frac{1\cdot \left(4-x\right)}{(4+x)(4-x)}-\frac{1\cdot \left(4+x\right)}{(4+x)(4-x)}=}$${\displaystyle \frac{\left(4-x\right)-\left(4+x\right)}{(4+x)(4-x)}=}$${\displaystyle \frac{\stackrel{1}{\cancel{4}}-x\cancel{-4}-x}{16-x^2}=}$${\displaystyle \frac{-2x}{16-x^2}}$.