840 Вправа 840Алгебра 8 клас Бевз 2025 · § 16 · стор. 157

840. Спрости вираз.

а) ${\displaystyle \frac{2+a^2}{\sqrt{2}-a}-\frac{2\sqrt{2}a}{\sqrt{2}-a}=}$${\displaystyle \frac{a^2-2\sqrt{2}a+2}{\sqrt{2}-a}}$=

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

б) ${\displaystyle a-4=}$${\displaystyle (\sqrt{a}-2)(\sqrt{a}+2)}$;

${\displaystyle a-3\sqrt{a}=}$${\displaystyle \sqrt{a}(\sqrt{a}-3)}$;

${\displaystyle \frac{a-4}{a-3\sqrt{a}}\cdot \frac{\sqrt{a}}{\sqrt{a}+2}-\frac{1}{\sqrt{a}-3}}$=

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

в)

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

2) ${\displaystyle \frac{2}{x(1-y)}:\frac{2}{y-1}=}$${\displaystyle \frac{\cancel{2}(y-1)}{x(1-y)\cdot \cancel{2}}=}$${\displaystyle -\frac{1}{x}}$;

г)

1) ${\displaystyle \frac{\sqrt{a}}{\sqrt{a}-\sqrt{b}}-\frac{\sqrt{a}}{\sqrt{a}+\sqrt{b}}=}$${\displaystyle \frac{a+\sqrt{ab}-a+\sqrt{ab}}{a-b}=}$${\displaystyle \frac{2\sqrt{ab}}{a-b}}$;

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

3) ${\displaystyle \frac{2\sqrt{a}}{\sqrt{a}+\sqrt{b}}+\frac{2\sqrt{b}}{\sqrt{a}+\sqrt{b}}=}$${\displaystyle \frac{2(\sqrt{a}+\sqrt{b})}{\sqrt{a}+\sqrt{b}}=}$${\displaystyle 2}$.