Bài 276: $\frac{1+2\sqrt{x}-x\sqrt{x}}{3-x-\sqrt{2-x}}=2(\frac{1+x\sqrt{x}}{1+x})$ (trích bài của bạn mamanhkhoi2000)
ĐK: $0 \leq x \leq 2$
$\dfrac{1+2\sqrt{x}-x\sqrt{x}}{3-x-\sqrt{2-x}}=\dfrac{2(\sqrt{x}+1)(x-\sqrt{x}+1)}{1+x}$
$\iff \dfrac{(\sqrt{x}+1)(-x+\sqrt{x}+1)}{3-x-\sqrt{2-x}}=\dfrac{2(\sqrt{x}+1)(x-\sqrt{x}+1)}{1+x}$
Vì $\sqrt{x}+1 >0$
$\iff \dfrac{-x+\sqrt{x}+1}{3-x-\sqrt{2-x}}=\dfrac{2(x-\sqrt{x}+1)}{1+x}=\dfrac{2(-x+\sqrt{x}+1)+2(x-\sqrt{x}+1)}{6-2x-2\sqrt{2-x}+1+x}=\dfrac{4}{[(2-x)-2\sqrt{2-x}+1]+4}=\dfrac{4}{(\sqrt{2-x}-1)^2+4} \leq 1$ (tính chất dãy tỉ số bằng nhau)
$\rightarrow \dfrac{2(x-\sqrt{x}+1)}{1+x} \leq 1$
$\iff 2(x-\sqrt{x}+1) \leq 1+x$
$\iff x-2\sqrt{x}+1 \leq 0$
$\iff (\sqrt{x}-1)^2 \leq 0$
$\iff \sqrt{x}=1 \iff x=1$
Vậy $x=1$