653 Вправа 653Алгебра 8 клас Тарасенкова 2025 · § 15 · стор. 193

653. Спростіть вираз:

1) ${\displaystyle \frac{1^{\mathrm{\setminus }\left(1-\sqrt{a}\right)}}{\sqrt{a}}+\frac{1^{\mathrm{\setminus }\sqrt{a}}}{1-\sqrt{a}}=}$${\displaystyle \frac{1-\sqrt{a}+\sqrt{a}}{\sqrt{a}(1-\sqrt{a})}=}$${\displaystyle \frac{1}{\sqrt{a}(1-\sqrt{a})}=}$${\displaystyle \frac{1}{\sqrt{a}-{\left(\sqrt{a}\right)}^2}=}$${\displaystyle \frac{1}{\sqrt{a}-a}}$;

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

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

4) ${\displaystyle \frac{1}{3\sqrt{m}+3\sqrt{n}}+\frac{2}{2\sqrt{m}+2\sqrt{n}}=}$${\displaystyle \frac{1}{3(\sqrt{m}+\sqrt{n})}+\frac{\stackrel{1}{\cancel{2}}}{\underset{1}{\cancel{2}}(\sqrt{m}+\sqrt{n})}=}$${\displaystyle \frac{1}{3(\sqrt{m}+\sqrt{n})}+\frac{1^{\mathrm{\setminus }3}}{\sqrt{m}+\sqrt{n}}=}$${\displaystyle \frac{1+3}{3(\sqrt{m}+\sqrt{n})}=}$${\displaystyle \frac{4}{3\left(\sqrt{m}+\sqrt{n}\right)}}$;

5) ${\displaystyle \frac{1}{\sqrt{a}+\sqrt{2}}-\frac{1}{2\sqrt{a}+2\sqrt{2}}=}$${\displaystyle \frac{1^{\mathrm{\setminus }2}}{\sqrt{a}+\sqrt{2}}-\frac{1}{2(\sqrt{a}+\sqrt{2})}=}$${\displaystyle \frac{2-1}{2(\sqrt{a}+\sqrt{2})}=}$${\displaystyle \frac{1}{2\left(\sqrt{a}+\sqrt{2}\right)}}$;

6) ${\displaystyle \frac{x^{\mathrm{\setminus }\left(\sqrt{x}-1\right)}}{\sqrt{x}+1}-\frac{x^{\mathrm{\setminus }\left(\sqrt{x}+1\right)}}{\sqrt{x}-1}=}$${\displaystyle \frac{x(\sqrt{x}-1)-x(\sqrt{x}+1)}{{\left(\sqrt{x}\right)}^2-1^2}=}$${\displaystyle \frac{x\sqrt{x}-x-x\sqrt{x}-x}{x-1}=}$${\displaystyle \frac{-2x}{x-1}}$.