Question

\(M=\left(1+\frac{\sqrt{a}}{a+1}\right):\left(\frac{1}{\sqrt{a}-1}-\frac{2\sqrt{a}}{a\sqrt{a}+\sqrt{a}-a-1}\right)\)\(v\text{ới }a\ge0;a\ne1\)

a) Rút gọn M.

b) Tìm a sao cho M>0.

Answer 1

Lời giải:

a)

\(M=\frac{a+1+\sqrt{a}}{a+1}:\left[\frac{1}{\sqrt{a}-1}-\frac{2\sqrt{a}}{(\sqrt{a}-1)(a+1)}\right]=\frac{a+1+\sqrt{a}}{a+1}:\frac{a+1-2\sqrt{a}}{(\sqrt{a}-1)(a+1)}\)

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

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

b) Để $M>0\Leftrightarrow \frac{a+1+\sqrt{a}}{\sqrt{a}-1}>0$

$\Leftrightarrow \sqrt{a}-1>0$ (do $a+1+\sqrt{a}>0$ với mọi $a\in$ ĐKXĐ)

$\Leftrightarrow a>1$

Vậy $a>1$ thì $M>0$