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

а)

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

б)

$$\frac{a-1}{\frac{1}{a+a^{1/2}+1}}:\frac{a^{1/2}+1}{a^{3/2}-1}+2a^{1/2} = (a-1)(a+a^{1/2}+1)\cdot \frac{a^{3/2}-1}{a^{1/2}+1}+2a^{1/2}.$$
Разложим разность кубов:
$$a^{3/2}-1=(a^{1/2}-1)(a+ a^{1/2}+1).$$
Тогда
$$(a-1)(a+a^{1/2}+1)\cdot \frac{(a^{1/2}-1)(a+a^{1/2}+1)}{a^{1/2}+1}+2a^{1/2} = (a^{1/2}-1)^2+2a^{1/2}=a+1.$$

в)

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

г)

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

Ответ: а) $$\frac{x-y}{\frac{1}{x^4}} \cdot \frac{x^2y^4(1+x^2y^2)}{x^2+y^2}$$; б) $$a+1$$; в) $$\frac{2a(a-b)}{a^2-b}$$; г) $$x-1$$.