Giải thích các bước giải:
$\sqrt{x+3}+4\sqrt{x}-2x=6-\sqrt{5-x}$
$\to \sqrt{x+3}+4\sqrt{x}-2x+\sqrt{5-x}-6=0$
$\to (\sqrt{x+3}-2)-2(x-2\sqrt{x}+1)+(\sqrt{5-x}-2)=0$
$\to \dfrac{x+3-4}{\sqrt{x+3}+2}-2(\sqrt{x}-1)^2+\dfrac{5-x-4}{\sqrt{5-x}+2}=0$
$\to \dfrac{x-1}{\sqrt{x+3}+2}-2(\dfrac{x-1}{\sqrt{x}+1})^2+\dfrac{1-x}{\sqrt{5-x}+2}=0$
$\to x=1$
Hoặc
$\to \dfrac{1}{\sqrt{x+3}+2}-2.\dfrac{x-1}{(\sqrt{x}+1)^2}-\dfrac{1}{\sqrt{5-x}+2}=0$
$\to \dfrac{1}{\sqrt{x+3}+2}=2.\dfrac{x-1}{(\sqrt{x}+1)^2}+\dfrac{1}{\sqrt{5-x}+2}$
Ta có : $0\le x\le 5$
$+)x>1\to \sqrt{x+3}+2>\sqrt{5-x}+2$
$\to \dfrac{1}{\sqrt{x+3}+2}<2.\dfrac{x-1}{(\sqrt{x}+1)^2}+\dfrac{1}{\sqrt{5-x}+2}$
$\to$Vô nghiệm
$+)x<1\to \sqrt{x+3}+2>\sqrt{5-x}+2\to \dfrac{1}{\sqrt{x+3}+2}>2.\dfrac{x-1}{(\sqrt{x}+1)^2}+\dfrac{1}{\sqrt{5-x}+2} $
$\to$Vô nghiệm
Vậy $x=1$
