Question

\(Q=\left(\dfrac{1}{a-\sqrt{a}}+\dfrac{1}{\sqrt{a}-1}\right):\dfrac{\sqrt{a}+1}{\sqrt{4a}}\)

Tìm a để Q < 1

Answer 1

Lời giải:

ĐK: $a>0; a\neq 1$

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

\(Q<1\Leftrightarrow Q-1<0\Leftrightarrow \frac{2}{\sqrt{a}-1}-1<0\)

\(\Leftrightarrow \frac{3-\sqrt{a}}{\sqrt{a}-1}<0\). Có 2 TH xảy ra:

TH1: $3-\sqrt{a}>0; \sqrt{a}-1<0$

$\Leftrightarrow 0\leq a< 1$. Kết hợp ĐKXĐ: suy ra $0< a< 1$

TH2: $3-\sqrt{a}<0; \sqrt{a}-1>0$

$\Rightarrow a>9$

Vậy...........