Упр.16.91 ГДЗ Мордкович 8 класс (Алгебра)
15.94
а)
$$\frac{a-16}{\sqrt a+3}\cdot\frac1{a+4\sqrt a}-\frac{\sqrt a+4}{a-3\sqrt a}$$
$$=\frac{(\sqrt a-4)(\sqrt a+4)}{(\sqrt a+3)\cdot\sqrt a(\sqrt a+4)}-\frac{\sqrt a+4}{\sqrt a(\sqrt a-3)}$$
$$=\frac{\sqrt a-4}{\sqrt a(\sqrt a+3)}-\frac{\sqrt a+4}{\sqrt a(\sqrt a-3)}$$
$$=\frac{(\sqrt a-4)(\sqrt a-3)-(\sqrt a+4)(\sqrt a+3)}{\sqrt a(\sqrt a+3)(\sqrt a-3)}$$
$$=\frac{a-7\sqrt a+12-a-7\sqrt a-12}{\sqrt a(a-9)}$$
$$=-\frac{14\sqrt a}{\sqrt a(a-9)}=-\frac{14}{a-9}=\frac{14}{9-a}.$$б)
$$\frac{1-2\sqrt b}{2\sqrt b+1}+\frac{b+3\sqrt b}{4b-1}:\frac{3+\sqrt b}{4\sqrt b+2}$$
$$=\frac{1-2\sqrt b}{2\sqrt b+1}+\frac{\sqrt b(\sqrt b+3)}{(2\sqrt b-1)(2\sqrt b+1)}\cdot\frac{2(2\sqrt b+1)}{3+\sqrt b}$$
$$=\frac{1-2\sqrt b}{2\sqrt b+1}+\frac{2\sqrt b}{2\sqrt b-1}$$
$$=\frac{(1-2\sqrt b)(2\sqrt b-1)+2\sqrt b(2\sqrt b+1)}{(2\sqrt b+1)(2\sqrt b-1)}$$
$$=\frac{6\sqrt b-1}{4b-1}.$$в)
$$\frac{9x}{2\sqrt x-\sqrt y}:\frac{12\sqrt{x^3}}{4x-y}\cdot\frac{4}{6x+3\sqrt{xy}}$$
$$=\frac{9x}{2\sqrt x-\sqrt y}\cdot\frac{(2\sqrt x-\sqrt y)(2\sqrt x+\sqrt y)}{12x\sqrt x}\cdot\frac{4}{3\sqrt x(2\sqrt x+\sqrt y)}$$
$$=\frac{9x\cdot4}{12x\sqrt x\cdot3\sqrt x}=\frac1x.$$г)
$$\frac{\sqrt{mn^3}}{\sqrt m-\sqrt n}\cdot\frac{m-n}{6n\sqrt m}:\frac{\sqrt{mn}+n}{6m}$$
$$=\frac{n\sqrt{mn}}{\sqrt m-\sqrt n}\cdot\frac{(\sqrt m-\sqrt n)(\sqrt m+\sqrt n)}{6n\sqrt m}\cdot\frac{6m}{\sqrt n(\sqrt m+\sqrt n)}$$
$$=\frac{\sqrt{mn}}{\sqrt n}=m.$$
Ответ
а) $$\frac{14}{9-a}$$; б) $$\frac{6\sqrt b-1}{4b-1}$$; в) $$\frac1x$$; г) $$m$$.
