ĐK: \(x\ge-1\)
PT \(\Leftrightarrow x^2+5x+7=7\sqrt{\left(x+1\right)\left(x^2-x+1\right)}\)
Đặt \(\sqrt{x+1}=a;\sqrt{x^2-x+1}=b\Rightarrow6a^2+b^2=x^2+5x+7\)
PT \(\Leftrightarrow6a^2+b^2=7ab\Leftrightarrow\left(a-b\right)\left(6a-b\right)=0\)
*Với a = b \(\Leftrightarrow\sqrt{x^2-x+1}=\sqrt{x+1}\Leftrightarrow x^2-x+1=x+1\)
\(\Leftrightarrow x\left(x-2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\left(TM\right)\\x=2\left(TM\right)\end{matrix}\right.\)
*Với \(6a=b\Leftrightarrow\sqrt{x^2-x+1}=6\sqrt{x+1}\)
\(\Leftrightarrow x^2-x+1=36x+36\)
\(\Leftrightarrow x^2-37x-35=0\) .Dùng delta tính nốt:v
Vậy..... (có 4 nghiệm thỏa mãn)...
