Упр.16.95 ГДЗ Мордкович 8 класс (Алгебра)

15.99

а)

$$ \frac{\dfrac{x}{x-\sqrt2}-\dfrac{\sqrt2}{x+\sqrt2}}{\dfrac{x^2+2}{x^2+x\sqrt2}} = \frac{\dfrac{x(x+\sqrt2)-\sqrt2(x-\sqrt2)}{(x-\sqrt2)(x+\sqrt2)}}{\dfrac{x^2+2}{x(x+\sqrt2)}} $$

$$ = \frac{\dfrac{x^2+x\sqrt2-x\sqrt2+2}{(x-\sqrt2)(x+\sqrt2)}}{\dfrac{x^2+2}{x(x+\sqrt2)}} = \frac{\dfrac{x^2+2}{(x-\sqrt2)(x+\sqrt2)}}{\dfrac{x^2+2}{x(x+\sqrt2)}} $$

$$ = \frac{x(x+\sqrt2)}{(x-\sqrt2)(x+\sqrt2)} = \frac{x}{x-\sqrt2} $$

б)

$$ \frac{\dfrac{\sqrt a}{\sqrt a-\sqrt b}-\dfrac{\sqrt b}{\sqrt a+\sqrt b}}{\dfrac{a^2+ab}{a-b}} = \frac{\dfrac{\sqrt a(\sqrt a+\sqrt b)-\sqrt b(\sqrt a-\sqrt b)}{(\sqrt a-\sqrt b)(\sqrt a+\sqrt b)}}{\dfrac{a(a+b)}{a-b}} $$

$$ = \frac{\dfrac{a+\sqrt{ab}-\sqrt{ab}+b}{a-b}}{\dfrac{a(a+b)}{a-b}} = \frac{\dfrac{a+b}{a-b}}{\dfrac{a(a+b)}{a-b}} = \frac{1}{a} $$

Ответ

а) $$\frac{x}{x-\sqrt2}$$;
б) $$\frac{1}{a}$$.