Đk: \(x\ne-3\) và \(x\ge-1\)
pt <=> \(\sqrt{\frac{x^3+1}{x+3}}-\sqrt{x+3}=\sqrt{x^2-x+1}-\sqrt{x-1}\)
=> \(\frac{x^3+1}{x+3}-2\sqrt{\frac{x^3+1}{x+3}.\left(x+3\right)}+x+3=x^2-x+1-2\sqrt{\left(x^2-x+1\right)\left(x+1\right)}+x+1\)
<=> \(\frac{x^3+1}{x+3}-2\sqrt{x^3+1}+x+3=x^2+2-2\sqrt{x^3+1}\)
<=> \(\frac{x^3+1}{x+3}+x+3-\left(x^2+2\right)=0\)
<=> \(\frac{x^3+1+\left(x+3\right)^2-\left(x+3\right)\left(x^2+2\right)}{x+3}=0\)
<=> \(x^3+1+x^2+6x+9-x^3-2x-3x^2-6=0\)
<=>\(4-2x^2+4x=0\)
<=> \(x^2-2x-2=0\)
<=> \(\left(x-1\right)^2=3\)
=> \(\left[{}\begin{matrix}x-1=-\sqrt{3}\\x-1=\sqrt{3}\end{matrix}\right.\) <=> \(\left[{}\begin{matrix}x=-\sqrt{3}+1\\x=\sqrt{3}+1\end{matrix}\right.\left(ktm\right)\)
Vậy pt vô nghiệm.
