Упр.16.46 ГДЗ Мордкович 8 класс (Алгебра)

1. $$\frac{1}{\sqrt{a+3}-2}=\frac{1}{\sqrt{a+3}-2}\cdot\frac{\sqrt{a+3}+2}{\sqrt{a+3}+2}$$

   $$=\frac{\sqrt{a+3}+2}{(\sqrt{a+3}-2)(\sqrt{a+3}+2)}=\frac{\sqrt{a+3}+2}{a+3-4}=\frac{\sqrt{a+3}+2}{a-1}.$$

2. $$\frac{y-3}{\sqrt{4-y}+1}=\frac{y-3}{\sqrt{4-y}+1}\cdot\frac{\sqrt{4-y}-1}{\sqrt{4-y}-1}$$

   $$=\frac{(y-3)(\sqrt{4-y}-1)}{(\sqrt{4-y}+1)(\sqrt{4-y}-1)}=\frac{(y-3)(\sqrt{4-y}-1)}{4-y-1}$$

   $$=\frac{(y-3)(\sqrt{4-y}-1)}{3-y}=\frac{-(3-y)(\sqrt{4-y}-1)}{3-y}=1-\sqrt{4-y}.$$

3. $$\frac{2}{3-\sqrt{2x-1}}=\frac{2}{3-\sqrt{2x-1}}\cdot\frac{3+\sqrt{2x-1}}{3+\sqrt{2x-1}}$$

   $$=\frac{2(3+\sqrt{2x-1})}{(3-\sqrt{2x-1})(3+\sqrt{2x-1})}=\frac{2(3+\sqrt{2x-1})}{9-(2x-1)}$$

   $$=\frac{2(3+\sqrt{2x-1})}{10-2x}=\frac{3+\sqrt{2x-1}}{5-x}.$$

4. $$\frac{3-b}{2-\sqrt{b+1}}=\frac{3-b}{2-\sqrt{b+1}}\cdot\frac{2+\sqrt{b+1}}{2+\sqrt{b+1}}$$

   $$=\frac{(3-b)(2+\sqrt{b+1})}{(2-\sqrt{b+1})(2+\sqrt{b+1})}=\frac{(3-b)(2+\sqrt{b+1})}{4-(b+1)}$$

   $$=\frac{(3-b)(2+\sqrt{b+1})}{3-b}=2+\sqrt{b+1}.$$

Ответ

а) $$\frac{\sqrt{a+3}+2}{a-1}$$;
б) $$1-\sqrt{4-y}$$;
в) $$\frac{3+\sqrt{2x-1}}{5-x}$$;
г) $$2+\sqrt{b+1}$$.