Question

Помогите решить и упростить выражение. Пожалуйста

Answer 1

$(\frac{\sqrt{a}+1}{\sqrt{a}-1}+(\frac{\sqrt{a}+1}{\sqrt{a}-1})^{-1}+4\sqrt{a})\cdot (\sqrt{a}-\frac{1}{\sqrt{a}})=\\\\=(\frac{\sqrt{a}+1}{\sqrt{a}-1}+\frac{\sqrt{a}-1}{\sqrt{a}+1}+4\sqrt{a})\cdot \frac{a-1}{\sqrt{a}}=\\\\=\frac{(\sqrt{a}+1)^2+(\sqrt{a}-1)^2+4\sqrt{a}(\sqrt{a}+1)(\sqrt{a}-1)}{(\sqrt{a}-1)(\sqrt{a}+1)}\cdot \frac{a-1}{\sqrt{a}}=\\\\=\frac{a+1+2\sqrt{a}+a+1-2\sqrt{a}+4\sqrt{a}(a-1)}{a-1}\cdot \frac{a-1}{\sqrt{a}}=\frac{2a+2+4a\sqrt{a}-4\sqrt{a}}{\sqrt{a}}=\\\\=\frac{2(a+1)}{\sqrt{a}}+4a-4=\frac{2\sqrt{a}(a+1)}{a}+4(a-1)$

Если перед второй дробью стоит минус, то

$b)\; \; (\frac{\sqrt{a}+1}{\sqrt{a}-1}-(\frac{\sqrt{a}+1}{\sqrt{a}-1})^{-1}+4\sqrt{a})\cdot (\sqrt{a}-\frac{1}{\sqrt{a}})=\\\\=(\frac{\sqrt{a}+1}{\sqrt{a}-1}-\frac{\sqrt{a}-1}{\sqrt{a}+1}+4\sqrt{a})\cdot \frac{a-1}{\sqrt{a}}=\\\\=\frac{(\sqrt{a}+1)^2-(\sqrt{a}-1)^2+4\sqrt{a}(\sqrt{a}+1)(\sqrt{a}-1)}{(\sqrt{a}-1)(\sqrt{a}+1)}\cdot \frac{a-1}{\sqrt{a}}=\\\\=\frac{a+2\sqrt{a}+1-a+2\sqrt{a}-1+4\sqrt{a}(a-1)}{a-1}\cdot \frac{a-1}{\sqrt{a}}=\frac{4\sqrt{a}+4\sqrt{a}(a-1)}{\sqrt{a}}=\\\\=\frac{4\sqrt{a}(1+a-1)}{\sqrt{a}}=\frac{4\sqrt{a}\cdot a}{\sqrt{a}}=4a$