Đk:\(x\ge0\)

\(\sqrt{x+3}+\sqrt{3x+1}=2\sqrt{x}+\sqrt{2x+2}\)

\(pt\Leftrightarrow\sqrt{x+3}-2+\sqrt{3x+1}-2=2\sqrt{x}-2+\sqrt{2x+2}-2\)

\(\Leftrightarrow\frac{x+3-4}{\sqrt{x+3}+2}+\frac{3x+1-4}{\sqrt{3x+1}-2}=\frac{4x-4}{2\sqrt{x}+2}+\frac{2x+2-4}{\sqrt{2x+2}+2}\)

\(\Leftrightarrow\frac{x-1}{\sqrt{x+3}+2}+\frac{3x-3}{\sqrt{3x+1}-2}=\frac{4x-4}{2\sqrt{x}+2}+\frac{2x-2}{\sqrt{2x+2}+2}\)

\(\Leftrightarrow\frac{x-1}{\sqrt{x+3}+2}+\frac{3\left(x-1\right)}{\sqrt{3x+1}-2}-\frac{4\left(x-1\right)}{2\sqrt{x}+2}-\frac{2\left(x-1\right)}{\sqrt{2x+2}+2}=0\)

\(\Leftrightarrow\left(x-1\right)\left(\frac{1}{\sqrt{x+3}+2}+\frac{3}{\sqrt{3x+1}-2}-\frac{4}{2\sqrt{x}+2}-\frac{2}{\sqrt{2x+2}+2}\right)=0\)

Dễ thấy: \(\frac{1}{\sqrt{x+3}+2}+\frac{3}{\sqrt{3x+1}-2}-\frac{4}{2\sqrt{x}+2}-\frac{2}{\sqrt{2x+2}+2}>0\)

\(\Rightarrow x-1=0\Rightarrow x=1\)

tự làm đi

\(x^2-x-4=2\sqrt{\left(x-1\right)\left(1-x\right)}\)

\(\Leftrightarrow\left(x^2-x-4\right)^2=\left[2\sqrt{\left(x-1\right)\left(1-x\right)}\right]^2\)

\(\Leftrightarrow x^4-2x^3-7x^2+8x+16=8x-4x^2-4\)

\(\Leftrightarrow x^4-2x^3-7x^2+8x+16+4=8x-4x^2\)

\(\Leftrightarrow x^4-2x^3-7x^2+8x+20=8x-4x^2\)

\(\Leftrightarrow x^4-2x^3-7x^2+8x+20+4x^2=8x\)

\(\Leftrightarrow x^4-2x^3-3x^2+8x+20=8x\)

\(\Leftrightarrow x^4-2x^3-3x^2+8x+20-8x=0\)

\(\Leftrightarrow x^4-2x^3-3x^2+20=0\)

Vậy: phương trình vô nghiệm

Có gì không hiểu ib