841 Вправа 841Алгебра 8 клас Бевз 2025 · § 16 · стор. 157

841. Спрости вираз.

а) ${\displaystyle \frac{a+3\sqrt{ab}}{\sqrt{a}+\sqrt{b}}-\frac{\sqrt{ab}-b}{\sqrt{a}+\sqrt{b}}-\sqrt{b}=}$${\displaystyle \frac{a+2\sqrt{ab}+b}{\sqrt{a}+\sqrt{b}}-\sqrt{b}}$=

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

б) ${\displaystyle 9a+6\sqrt{a}+1=}$${\displaystyle {(3\sqrt{a}+1)}^2}$;

1) ${\displaystyle \frac{3\sqrt{a}-6}{9a+6\sqrt{a}+1}:\frac{\sqrt{a}-2}{3\sqrt{a}+1}=}$${\displaystyle \frac{3\cancel{\left(\sqrt{a}-2\right)}(3\sqrt{a}+1)}{{(3\sqrt{a}+1)}^2\cancel{\left(\sqrt{a}-2\right)}}=}$${\displaystyle \frac{3}{3\sqrt{a}+1}}$;

2) ${\displaystyle \frac{3}{3\sqrt{a}+1}+\frac{9\sqrt{a}}{3\sqrt{a}+1}=}$${\displaystyle \frac{3(1+3\sqrt{a})}{3\sqrt{a}+1}=}$${\displaystyle 3}$;

в)

1) ${\displaystyle \frac{\sqrt{x}}{\sqrt{x}+\sqrt{y}}+\frac{\sqrt{x}}{\sqrt{x}-\sqrt{y}}=}$${\displaystyle \frac{x-\sqrt{xy}+x+\sqrt{xy}}{x-y}=}$${\displaystyle \frac{2x}{x-y}}$;

2) ${\displaystyle \frac{2x}{x-y}\cdot \frac{x-2\sqrt{xy}+y}{x}}$=

=${\displaystyle \frac{2\cancel{x}{(\sqrt{x}-\sqrt{y})}^2}{\cancel{x}(\sqrt{x}-\sqrt{y})(\sqrt{x}+\sqrt{y})}=}$${\displaystyle \frac{2(\sqrt{x}-\sqrt{y})}{\sqrt{x}+\sqrt{y}}}$;

3) ${\displaystyle \frac{2\sqrt{x}-2\sqrt{y}}{\sqrt{x}+\sqrt{y}}+\frac{4\sqrt{y}}{\sqrt{x}+\sqrt{y}}=}$${\displaystyle \frac{2\sqrt{x}+2\sqrt{y}}{\sqrt{x}+\sqrt{y}}=}$${\displaystyle 2}$.