Đáp án : $x=1$
Giải thích các bước giải:
Ta có :
$\dfrac{1}{\sqrt{x+3}}+\dfrac{1}{\sqrt{3x+1}}=\dfrac{2}{1+\sqrt{x}}$
$\to (\dfrac{1}{\sqrt{x+3}}-\dfrac{1}{1+\sqrt{x}})+(\dfrac{1}{\sqrt{3x+1}}-\dfrac{1}{1+\sqrt{x}})=0$
$\to \dfrac{1+\sqrt{x}-\sqrt{x+3}}{\sqrt{x+3}(1+\sqrt{x})}+\dfrac{1+\sqrt{x}-\sqrt{3x+1}}{\sqrt{3x+1}(1+\sqrt{x})}=0$
$\to \dfrac{(\sqrt{x}-1)-(\sqrt{x+3}-2)}{\sqrt{x+3}(1+\sqrt{x})}+\dfrac{(\sqrt{x}-1)-(\sqrt{3x+1}-2)}{\sqrt{3x+1}(1+\sqrt{x})}=0$
$\to \dfrac{\dfrac{x-1}{\sqrt{x}+1}-\dfrac{x+3-4}{\sqrt{x+3}+2}}{\sqrt{x+3}(1+\sqrt{x})}+\dfrac{\dfrac{x-1}{\sqrt{x}+1}-\dfrac{3x+1-4}{\sqrt{3x+1}+2}}{\sqrt{3x+1}(1+\sqrt{x})}=0$
$\to \dfrac{\dfrac{x-1}{\sqrt{x}+1}-\dfrac{x-1}{\sqrt{x+3}+2}}{\sqrt{x+3}(1+\sqrt{x})}+\dfrac{\dfrac{x-1}{\sqrt{x}+1}-\dfrac{3(x-1)}{\sqrt{3x+1}+2}}{\sqrt{3x+1}(1+\sqrt{x})}=0$
$\to x-1=0\to x=1$
Hoặc
$\to \dfrac{\dfrac{1}{\sqrt{x}+1}-\dfrac{1}{\sqrt{x+3}+2}}{\sqrt{x+3}(1+\sqrt{x})}+\dfrac{\dfrac{1}{\sqrt{x}+1}-\dfrac{3}{\sqrt{3x+1}+2}}{\sqrt{3x+1}(1+\sqrt{x})}=0$
Vì $x\ge 0\to \sqrt{x}+1< \sqrt{x+3}+2\to \dfrac{1}{\sqrt{x}+1}-\dfrac{1}{\sqrt{x+3}+2}>0$
$\sqrt{x}+1<\dfrac{\sqrt{3x+1}+2}{3}$ luôn đúng
$\to \dfrac{\dfrac{1}{\sqrt{x}+1}-\dfrac{1}{\sqrt{x+3}+2}}{\sqrt{x+3}(1+\sqrt{x})}+\dfrac{\dfrac{1}{\sqrt{x}+1}-\dfrac{3}{\sqrt{3x+1}+2}}{\sqrt{3x+1}(1+\sqrt{x})}>0$
$\to$Phương trình vô nghiệm
