Đặt \(\sqrt{x+1}=a\) \(ĐKXĐ:x\ge0\)
\(\sqrt{3x}=b\)
Ta có: \(a-b=b^2-a^2\)
\(\Leftrightarrow a-b+a^2-b^2=0\)
\(\Leftrightarrow\left(a-b\right)+\left(a+b\right)\left(a-b\right)=0\)
\(\Leftrightarrow\left(a-b\right)\left(a+b+1\right)=0\)
Mà \(a+b+1>0\forall x\)
\(\Rightarrow a-b=0\)
\(\Leftrightarrow a=b\)
\(\Leftrightarrow\sqrt{x+1}=\sqrt{3x}\)
\(\Leftrightarrow x+1=3x\)
\(\Leftrightarrow2x-1=0\)
\(\Leftrightarrow x=\frac{1}{2}\)
Vậy phương trình có tập nghiệm \(S=\left\{\frac{1}{2}\right\}\)
\(ĐKXĐ:x\ge0\)
Ta có PT \(\Leftrightarrow\sqrt{x+1}-\sqrt{3x}-\left(2x-1\right)=0\)
\(\Leftrightarrow\left(\sqrt{x+1}-\frac{\sqrt{6}}{2}\right)-\left(\sqrt{3x}-\frac{\sqrt{6}}{2}\right)-\left(2x-1\right)=0\)
\(\Leftrightarrow\frac{x+1-\frac{6}{4}}{\sqrt{x+1}+\frac{\sqrt{6}}{2}}-\frac{3x-\frac{6}{4}}{\sqrt{3x}+\frac{\sqrt{6}}{2}}-\left(2x-1\right)=0\)
\(\Leftrightarrow\frac{x-\frac{1}{2}}{\sqrt{x+1}+\frac{\sqrt{6}}{2}}-\frac{3\left(x-\frac{1}{2}\right)}{\sqrt{3x}+\frac{\sqrt{6}}{2}}-2\left(x-\frac{1}{2}\right)=0\)
\(\Leftrightarrow\left(x-\frac{1}{2}\right)\left(\frac{1}{\sqrt{x+1}+\frac{\sqrt{6}}{2}}-\frac{3}{\sqrt{3x}+\frac{\sqrt{6}}{2}}-2\right)=0\)
\(\Rightarrow x=\frac{1}{2}\)(TMĐKXĐ)