Đk: \(x\ge-1\)
pt<=> \(3\left(x^2+2x+2\right)=10\sqrt{\left(x+1\right)\left(x^2-x+1\right)+2x\left(x+1\right)}\)
\(3\left(x^2+2x+2\right)=10\sqrt{\left(x+1\right)\left(x^2-x+2x+1\right)}\)
<=> \(3\left(x^2+2x+1\right)=10\sqrt{\left(x+1\right)\left(x^2+x+1\right)}\)
Đặt \(\sqrt{x+1}=a\left(a\ge0\right)\),\(\sqrt{x^2+x+1}=b\)
=> \(a^2+b^2=x+1+x^2+x+1=x^2+2x+2\)
Có \(3\left(a^2+b^2\right)=10ab\)
<=>\(3a^2-10ab+3b^2=0\)
<=> \(3a^2-ab-9ab+3b^2=0\)
<=> \(a\left(3a-b\right)-3b\left(3a-b\right)=0\)
<=> \(\left(a-3b\right)\left(3a-b\right)=0\) <=> \(\left[{}\begin{matrix}a=3b\\3a=b\end{matrix}\right.\)
<=>\(\left[{}\begin{matrix}\sqrt{x+1}=3\sqrt{x^2+x+1}\\3\sqrt{x+1}=\sqrt{x^2+x+1}\end{matrix}\right.\)
Giải nốt :))
