Упр.16.79 ГДЗ Мордкович 8 класс (Алгебра)

1. а)
   $$\left(2+\frac{\sqrt t}{\sqrt t+1}\right)\cdot \frac{3t+3\sqrt t}{12\sqrt t+8}$$
   $$=\frac{2\sqrt t+2+\sqrt t}{\sqrt t+1}\cdot \frac{3\sqrt t(\sqrt t+1)}{4(3\sqrt t+2)}$$
   $$=\frac{3\sqrt t+2}{\sqrt t+1}\cdot \frac{3\sqrt t(\sqrt t+1)}{4(3\sqrt t+2)}=\frac{3\sqrt t}{4}.$$

2. б)
   $$\left(\frac{\sqrt x-2\sqrt y}{\sqrt{xy}}+\frac1{\sqrt x}\right)\cdot \frac{xy}{\sqrt x-\sqrt y}$$
   $$=\frac{\sqrt x-2\sqrt y+\sqrt y}{\sqrt{xy}}\cdot \frac{xy}{\sqrt x-\sqrt y}$$
   $$=\frac{\sqrt x-\sqrt y}{\sqrt{xy}}\cdot \frac{xy}{\sqrt x-\sqrt y}=\sqrt{xy}.$$

3. в)
   $$\left(\sqrt a-\frac{a}{\sqrt a+1}\right)\cdot \frac{a-1}{\sqrt a}$$
   $$=\frac{\sqrt a(\sqrt a+1)-a}{\sqrt a+1}\cdot \frac{a-1}{\sqrt a}$$
   $$=\frac{a+\sqrt a-a}{\sqrt a+1}\cdot \frac{(\sqrt a-1)(\sqrt a+1)}{\sqrt a}$$
   $$=\frac{\sqrt a(\sqrt a-1)}{\sqrt a}=\sqrt a-1.$$

4. г)
   $$\frac{\sqrt{cd}-d}{c+d}\cdot \left(\frac{\sqrt c}{\sqrt c+\sqrt d}+\frac{\sqrt d}{\sqrt c-\sqrt d}\right)$$
   $$=\frac{\sqrt{cd}-d}{c+d}\cdot \frac{\sqrt c(\sqrt c-\sqrt d)+\sqrt d(\sqrt c+\sqrt d)}{(\sqrt c+\sqrt d)(\sqrt c-\sqrt d)}$$
   $$=\frac{\sqrt{cd}-d}{c+d}\cdot \frac{c-\sqrt{cd}+\sqrt{cd}+d}{(\sqrt c+\sqrt d)(\sqrt c-\sqrt d)}$$
   $$=\frac{\sqrt{cd}-d}{c+d}\cdot \frac{c+d}{(\sqrt c+\sqrt d)(\sqrt c-\sqrt d)}$$
   $$=\frac{\sqrt d(\sqrt c-\sqrt d)}{(\sqrt c+\sqrt d)(\sqrt c-\sqrt d)}=\frac{\sqrt d}{\sqrt c+\sqrt d}.$$

Ответ

а) $$\frac{3\sqrt t}{4}$$;
б) $$\sqrt{xy}$$;
в) $$\sqrt a-1$$;
г) $$\frac{\sqrt d}{\sqrt c+\sqrt d}$$.