Упр.435 ГДЗ Колмогоров 10-11 класс (Алгебра)

Подробный ответ

а)

$$\frac{x-y}{\frac{3}{x^4}+\frac{1}{x^2y^4}} :\frac{\frac{1}{x^2y^4}+\frac{1}{x^4y^2}}{\frac{1}{x^2+y^2}} = \frac{x-y}{\frac{3y^4+x^2}{x^4y^4}} :\frac{\frac{x^2+y^2}{x^4y^4}}{\frac{1}{x^2+y^2}}$$
$$= \frac{x-y}{\frac{3y^4+x^2}{x^4y^4}} :\frac{(x^2+y^2)^2}{x^4y^4} = (x-y)\cdot \frac{x^4y^4}{3y^4+x^2}\cdot \frac{x^4y^4}{(x^2+y^2)^2}$$

После сокращения получаем:
$$\frac{\left(\frac{1}{x^2}-\frac{1}{y^2}\right)\frac{1}{y^4}}{\frac{1}{x^4}} = \frac{x^4}{y^4}\left(\frac{1}{x^2}-\frac{1}{y^2}\right).$$

б)

$$\frac{a-1}{\frac{1}{a+a^{1/2}+1}}:\frac{a^{1/2}+1}{a^{3/2}-1}+2a^{1/2}$$
$$= (a-1)\cdot \frac{a+a^{1/2}+1}{a^{1/2}+1}\cdot \frac{a^{3/2}-1}{a^{1/2}+1}+2a^{1/2}$$
$$= \left(a^{1/2}-1\right)^2+2a^{1/2} = a+1.$$

в)

$$\left(\frac{1}{a+a^{1/2}b^{1/2}}+\frac{1}{a-a^{1/2}b^{1/2}}\right)\cdot \frac{a^3-b^3}{a^2+ab+b^2}$$
$$= \left(\frac{1}{a^{1/2}\left(a^{1/2}+b^{1/2}\right)}+\frac{1}{a^{1/2}\left(a^{1/2}-b^{1/2}\right)}\right)\cdot \frac{(a-b)(a^2+ab+b^2)}{a^2+ab+b^2}$$
$$= \frac{2a^{1/2}}{a^{1/2}(a-b)}\cdot (a-b)=2.$$

г)

$$\frac{\sqrt{x}+1}{x\sqrt{x}+x+\sqrt{x}}:\frac{1}{x^2-\sqrt{x}} = \frac{\sqrt{x}+1}{\sqrt{x}(\,x+\sqrt{x}+1\,)}\cdot (x^2-\sqrt{x})$$
$$= \frac{\sqrt{x}+1}{\sqrt{x}(\,x+\sqrt{x}+1\,)}\cdot \sqrt{x}(x\sqrt{x}-1)$$
$$= \frac{(\sqrt{x}+1)(x\sqrt{x}-1)}{x+\sqrt{x}+1} = x-1.$$