Đk: \(x\ge-1\)
pt <=>\(5\left[\sqrt{x^3+1}-2\left(x+1\right)\right]=2x^2+4-10\left(x+1\right)\)
Với \(x\ge-1\)=> \(\sqrt{x^3+1}+2\left(x+1\right)\ge0\)
pt <=> \(5.\frac{x^3+1-4\left(x+1\right)^2}{\sqrt{x^3+1}+2\left(x+1\right)}=2x^2-10x-6\)
<=> \(5.\frac{x^3-4x^2-8x-3}{\sqrt{x^3+1}+2\left(x+1\right)}-2\left(x^2-5x-3\right)=0\)
<=>\(\frac{5\left(x+1\right)\left(x^2-5x-3\right)}{\sqrt{x^3+1}+2\left(x+1\right)}-2\left(x^2-5x-3\right)=0\)
<=>\(\left(x^2-5x-3\right)\left(\frac{5\left(x+1\right)}{\sqrt{x^3+1}+2\left(x+1\right)}-2\right)=0\)
<=> \(\left[{}\begin{matrix}x^2-5x-3=0\left(1\right)\\5\left(x+1\right)=2\sqrt{x^3+1}+4\left(x+1\right)\left(2\right)\end{matrix}\right.\)
pt (1) <=> \(\left[{}\begin{matrix}x=\frac{5-\sqrt{37}}{2}\\x=\frac{5+\sqrt{37}}{2}\end{matrix}\right.\)(thỏa mãn)
pt (2) <=> \(x+1=2\sqrt{x^3+1}\)
<=> \(x^2+2x+1-4x^3-4=0\)
<=>\(-4x^3+x^2+2x-3=0\) <=> \(\left(x+1\right)\left(-4x^2+5x-3\right)=0\)
<=> \(\left(x+1\right)\left[-\left(2x-\frac{5}{4}\right)^2-\frac{23}{16}\right]=0\)
<=>x+1=0 (do \(-\left(2x-\frac{5}{4}\right)^2-\frac{23}{16}< 0\forall x\))
<=>x=-1(ktm)
Vậy pt có tập nghiệm \(S=\left\{\frac{5-\sqrt{37}}{2},\frac{5+\sqrt{37}}{2}\right\}\)
