cho A= [tex]1+(\dfrac{2a+ \sqrt{a}-1}{1-a}-\dfrac{2a\sqrt{a}-\sqrt{a}+a}{1-a\sqrt{a}}).\dfrac{a-\sqrt{a}}{2\sqrt{a}-1}[/tex]
a) Tìm ĐKXĐ của A
b) Rút gọn A
c) chứng minh A> [tex]\dfrac{2}{3}[/tex]
a) ĐKXĐ: $a\ge 0;a\ne \dfrac14;a\ne 1$
b) $A=1+\left( \dfrac{2a+\sqrt a-1}{1-a}-\dfrac{2a\sqrt a-\sqrt a+a}{1-a\sqrt a} \right).\dfrac{a-\sqrt a}{2\sqrt a-1}$
$=1+\left[ \dfrac{(\sqrt a+1)(2\sqrt a-1)}{(\sqrt a+1)(1-\sqrt a)}-\dfrac{\sqrt a(\sqrt a+1)(2\sqrt a-1)}{(1-\sqrt a)(a+\sqrt a+1)} \right].\dfrac{\sqrt a(\sqrt a-1)}{2\sqrt a-1}$
$=1+\dfrac{(2\sqrt a-1)(a+\sqrt a+1)-(a+\sqrt a)(2\sqrt a-1)}{(1-\sqrt a)(a+\sqrt a+1)}.\dfrac{\sqrt a(\sqrt a-1)}{2\sqrt a-1}$
$=1-\dfrac{(2\sqrt a-1)(a+\sqrt a+1-a-\sqrt a)}{(\sqrt a-1)(a+\sqrt a+1)}.\dfrac{\sqrt a(\sqrt a-1)}{2\sqrt a-1}$
$=1-\dfrac{\sqrt a}{a+\sqrt a+1}=\dfrac{a+\sqrt a+1-\sqrt a}{a+\sqrt a+1}=\dfrac{a+1}{a+\sqrt a+1}$
c) $A=\dfrac{a+1}{a+\sqrt a+1}\ge \dfrac{a+1}{a+1+\dfrac{a+1}2}=\dfrac{a+1}{\dfrac 32(a+1)}=\dfrac 23$ (AM-GM)
Dấu '=' xảy ra khi $a=1$ (KTM) => Dấu '=' không xảy ra $\Rightarrow A>\dfrac 23$ (đpcm)
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