Упр.10.27 ГДЗ Мерзляк 10 класс Базовый уровень (Алгебра)

1. Докажите тождество:
   1. $$\left(\frac{a^{0,5}+2}{a+2a^{0,5}+1}-\frac{a^{0,5}-2}{a-1}\right):\frac{a^{0,5}}{a^{0,5}+1}=\frac{2}{a-1};$$
   2. $$\frac{(a-b)^2}{a^{3/2}-b^{3/2}}+\frac{a^2-b^2}{\left(a^{1/2}+b^{1/2}\right)\left(a+a^{1/2}b^{1/2}+b\right)}=2a^{1/2}-2b^{1/2}.$$

1)

$$\left(\frac{\sqrt a+2}{a+2\sqrt a+1}-\frac{\sqrt a-2}{a-1}\right):\frac{\sqrt a}{\sqrt a+1}$$
$$=\left(\frac{\sqrt a+2}{(\sqrt a+1)^2}-\frac{\sqrt a-2}{(\sqrt a-1)(\sqrt a+1)}\right)\cdot \frac{\sqrt a+1}{\sqrt a}$$
$$=\left(\frac{\sqrt a+2}{\sqrt a+1}-\frac{\sqrt a-2}{\sqrt a-1}\right)\cdot \frac{1}{\sqrt a}$$
$$=\frac{(\sqrt a+2)(\sqrt a-1)-(\sqrt a-2)(\sqrt a+1)}{(\sqrt a+1)(\sqrt a-1)}\cdot \frac{1}{\sqrt a}$$
$$=\frac{(a-\sqrt a+2\sqrt a-2)-(a+\sqrt a-2\sqrt a-2)}{(a-1)\sqrt a}$$
$$=\frac{2\sqrt a}{(a-1)\sqrt a} =\frac{2}{a-1}.$$

Тождество доказано.

2)

$$\frac{(a-b)^2}{a^{3/2}-b^{3/2}}+\frac{a^2-b^2}{(\sqrt a+\sqrt b)(a+\sqrt a\sqrt b+b)}$$
$$=\frac{(a-b)(a-b)}{(\sqrt a-\sqrt b)(a+\sqrt a\sqrt b+b)}+\frac{(a-b)(a+b)}{(\sqrt a+\sqrt b)(a+\sqrt a\sqrt b+b)}$$
$$=\frac{(\sqrt a-\sqrt b)(\sqrt a+\sqrt b)(a-b)}{(\sqrt a-\sqrt b)(a+\sqrt a\sqrt b+b)} +\frac{(\sqrt a-\sqrt b)(\sqrt a+\sqrt b)(a+b)}{(\sqrt a+\sqrt b)(a+\sqrt a\sqrt b+b)}$$
$$=\frac{(\sqrt a+\sqrt b)(a-b)+(\sqrt a-\sqrt b)(a+b)}{a+\sqrt a\sqrt b+b}$$
$$=\frac{a\sqrt a-a\sqrt b-b\sqrt a+b\sqrt b+a\sqrt a+a\sqrt b-b\sqrt a-b\sqrt b}{a+\sqrt a\sqrt b+b}$$
$$=\frac{2a\sqrt a-2b\sqrt b}{a+\sqrt a\sqrt b+b} =2\sqrt a-2\sqrt b.$$

Тождество доказано.