View attachment 28455
Mọi người giúp em vs ạ !! EM CẦN GẤP
))
a) ĐK: $x>0;x\ne 1$
$G=\left[ \dfrac{(1-\sqrt x)(1+\sqrt x+x)}{\sqrt x(1-\sqrt x)}+1 \right]\left[ \dfrac{(1+\sqrt x)(1-\sqrt x+x)}{1+\sqrt x} \right]: \dfrac{(1-x)^3}{1+\sqrt x}
\\=\left( \dfrac{1+\sqrt x+x}{\sqrt x}+1 \right).(1-\sqrt x+x-\sqrt x): \dfrac{(1+\sqrt x)^3(1-\sqrt x)^3}{1+\sqrt x}
\\=\dfrac{1+2\sqrt x+x}{\sqrt x}.(1-2\sqrt x+x):[(1+\sqrt x)^2(1-\sqrt x)^3]
\\=\dfrac{(1+\sqrt x)^2(1-\sqrt x)^2}{\sqrt x}.\dfrac1{(1+\sqrt x)^2(1-\sqrt x)^3}
\\=\dfrac1{\sqrt x(1-\sqrt x)}=\dfrac1{\sqrt x-x}$
b) $\sqrt G$ xác định $\Leftrightarrow G>0\Leftrightarrow \sqrt x-x>0\Leftrightarrow \sqrt x(1-\sqrt x)>0\Leftrightarrow \left\{\begin{matrix} \sqrt x>0 \\ 1-\sqrt x>0 \end{matrix} \right.\Leftrightarrow 0<x<1$
c) Khi $x=3-2\sqrt 2=(\sqrt 2-1)^2$ (TMĐK) $\Rightarrow \sqrt x=\sqrt 2-1$, thì giá trị của $G$ là:
$G=\dfrac1{\sqrt 2-1-3+2\sqrt 2}=\dfrac1{3\sqrt 2-4}=\dfrac{4+3\sqrt 2}2$