Question

[bùi kim anh]

\(A=\left\{\dfrac{a-\sqrt{a}}{\sqrt{a}-1}-\dfrac{\sqrt{a}+1}{a+\sqrt{a}}\right\}:\dfrac{\sqrt{a}+1}{a\sqrt{a}}\)  rút gọn bài tập sau!

Answer 1

[Akai Haruma]

Lời giải:
ĐKXĐ: $a>0; a\neq 1$

\(A=\left[\frac{\sqrt{a}(\sqrt{a}-1)}{\sqrt{a}-1}-\frac{\sqrt{a}+1}{\sqrt{a}(\sqrt{a}+1)}\right]:\frac{\sqrt{a}+1}{a\sqrt{a}}\)

\(=(\sqrt{a}-\frac{1}{\sqrt{a}}).\frac{a\sqrt{a}}{\sqrt{a}+1}=\frac{a-1}{\sqrt{a}}.\frac{a\sqrt{a}}{\sqrt{a}+1}=\frac{(\sqrt{a}-1)(\sqrt{a}+1)}{\sqrt{a}}.\frac{a\sqrt{a}}{\sqrt{a}+1}=a(\sqrt{a}-1)\)