Упр.950 ГДЗ Макарычев Миндюк 9 класс (Углубленный) (Алгебра)

а) (((a^(1/2)+b^(1/2))/(a^(3/2)+b^(3/2)))^(-1)-(a^(1/2)-b^(1/2))^2)·(vb)^(-1);
б) ((2+x^(1/4))/(2-x^(1/4))-(2-x^(1/4))/(2+x^(1/4)))·(4-vx)/(x^3)^(1/4);
в) ((x^(3/2)-a)/(x^(1/2)-a^(1/3))+a^(1/3) x^(1/2))·(x+a^(1/3) x^(1/2))^(-1)-a^(1/3) x^(-1/2);
г) ((a^0,5+b^0,5)/(a+b)^0,5-(a+b)^0,5/(a^0,5+b^0,5))^(-2)-(a+b)/(2v(ab)).

1. а)

   $$\left(\left(\frac{\sqrt a+\sqrt b}{a^{3/2}+b^{3/2}}\right)^{-1}-(\sqrt a-\sqrt b)^2\right)\cdot(\sqrt b)^{-1}$$
   $$=\left(\frac{a^{3/2}+b^{3/2}}{\sqrt a+\sqrt b}-(a-2\sqrt{ab}+b)\right)\cdot\frac1{\sqrt b}$$
   $$=\left(a-\sqrt{ab}+b-a+2\sqrt{ab}-b\right)\cdot\frac1{\sqrt b}$$
   $$=\sqrt a.$$

2. б)

   $$\left(\frac{2+x^{1/4}}{2-x^{1/4}}-\frac{2-x^{1/4}}{2+x^{1/4}}\right)\cdot\frac{4-\sqrt x}{\sqrt[4]{x^3}}$$
   $$=\frac{(2+x^{1/4})^2-(2-x^{1/4})^2}{(2-x^{1/4})(2+x^{1/4})}\cdot\frac{4-\sqrt x}{\sqrt[4]{x^3}}$$
   $$=\frac{8x^{1/4}}{4-\sqrt x}\cdot\frac{4-\sqrt x}{\sqrt[4]{x^3}}$$
   $$=\frac{8x^{1/4}}{x^{3/4}}=\frac{8}{x^{1/2}}=\frac{8}{\sqrt x}.$$

3. в)

   $$\left(\frac{x^{3/2}-a}{x^{1/2}-a^{1/3}}+a^{1/3}x^{1/2}\right)\cdot\left(x+a^{1/3}x^{1/2}\right)^{-1}-a^{1/3}x^{-1/2}$$
   $$=\left(\frac{x^{3/2}-a+a^{1/3}x^{1/2}(x^{1/2}-a^{1/3})}{x^{1/2}-a^{1/3}}\right)\cdot\frac1{x+a^{1/3}x^{1/2}}-a^{1/3}x^{-1/2}$$
   $$=\frac{x\left(x^{1/2}+a^{1/3}\right)-a^{2/3}\left(x^{1/2}+a^{1/3}\right)}{x^{1/2}\left(x^{1/2}-a^{1/3}\right)\left(x^{1/2}+a^{1/3}\right)}-a^{1/3}x^{-1/2}$$
   $$=\frac{x^{1/2}-a^{1/3}}{x^{1/2}\left(x^{1/2}-a^{1/3}\right)}-a^{1/3}x^{-1/2}=1.$$

4. г)

   $$\left(\frac{\sqrt a+\sqrt b}{\sqrt{a+b}}-\frac{\sqrt{a+b}}{\sqrt a+\sqrt b}\right)^{-2}-\frac{a+b}{2\sqrt{ab}}$$
   $$=\left(\frac{(\sqrt a+\sqrt b)^2-(a+b)}{\sqrt{a+b}(\sqrt a+\sqrt b)}\right)^{-2}-\frac{a+b}{2\sqrt{ab}}$$
   $$=\left(\frac{2\sqrt{ab}}{\sqrt{a+b}(\sqrt a+\sqrt b)}\right)^{-2}-\frac{a+b}{2\sqrt{ab}}$$
   $$=\left(\frac{\sqrt{a+b}(\sqrt a+\sqrt b)}{2\sqrt{ab}}\right)^2-\frac{a+b}{2\sqrt{ab}}$$
   $$=\frac{(a+b)(a+2\sqrt{ab}+b)}{4ab}-\frac{2\sqrt{ab}(a+b)}{4ab}$$
   $$=\frac{(a+b)(\sqrt a+\sqrt b)^2-2\sqrt{ab}(a+b)}{4ab}=\frac{(a+b)^2}{4ab}.$$

Ответ

а) $$\sqrt a$$; б) $$\frac{8}{\sqrt x}$$; в) $$1$$; г) $$\frac{(a+b)^2}{4ab}$$.