Упр.19.19 Часть 1 ГДЗ Мордкович Семенов 9 класс (Алгебра)

а) (x+v(x^2-xy))/(x-v(x^2-xy))+(x-v(x^2-xy))/(x+v(x^2-xy));
б) ((1+v(1-x^2))/(1-v(1-x^2))+1):(1+v(1-x^2)).

а)

$$ \frac{x+\sqrt{x^2-xy}}{x-\sqrt{x^2-xy}}+\frac{x-\sqrt{x^2-xy}}{x+\sqrt{x^2-xy}} = \frac{\left(x+\sqrt{x^2-xy}\right)^2+\left(x-\sqrt{x^2-xy}\right)^2}{\left(x-\sqrt{x^2-xy}\right)\left(x+\sqrt{x^2-xy}\right)} $$

$$ =\frac{x^2+2x\sqrt{x^2-xy}+x^2-xy+x^2-2x\sqrt{x^2-xy}+x^2-xy}{x^2-(x^2-xy)} $$

$$ =\frac{4x^2-2xy}{xy}=\frac{2(2x-y)}{y} $$

б)

$$ \left(\frac{1+\sqrt{1-x^2}}{1-\sqrt{1-x^2}}+1\right):\left(1+\sqrt{1-x^2}\right) $$

$$ =\frac{1+\sqrt{1-x^2}+1-\sqrt{1-x^2}}{1-\sqrt{1-x^2}} \cdot \frac{1}{1+\sqrt{1-x^2}} $$

$$ =\frac{2}{1-\left(1-x^2\right)}\cdot \frac{1}{1+\sqrt{1-x^2}} =\frac{2}{x^2}\cdot \frac{1}{1+\sqrt{1-x^2}} $$

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

Ответ

а) $$\frac{2(2x-y)}{y}$$; б) $$\frac{2}{x^2\left(1+\sqrt{1-x^2}\right)}$$.