\(2\sqrt{x+3}=9x^2-x-4\)
ĐKXĐ: \(x\ge-3\)
\(\Leftrightarrow2\sqrt{x+3}-4=9x^2-x-8\)
\(\Leftrightarrow\frac{2\left(x-1\right)}{\sqrt{x+3}+2}=\left(x-1\right)\left(9x+8\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\\frac{2}{\sqrt{x+3}+2}=9x+8\left(1\right)\end{matrix}\right.\)
Xét (1), đặt \(\sqrt{x+3}=t\ge0\Rightarrow x=t^2-3\)
\(\left(1\right)\Leftrightarrow\frac{2}{t+2}=9\left(t^2-3\right)+8\)
\(\Leftrightarrow\left(t+2\right)\left(9t^2-19\right)-2=0\)
\(\Leftrightarrow9t^3+18t^2-19t-40=0\)
\(\Leftrightarrow\left(3t+5\right)\left(3t^2+t-8\right)=0\)
\(\Leftrightarrow3t^2+t-8=0\Rightarrow\left[{}\begin{matrix}t=\frac{\sqrt{97}-1}{6}\\t=\frac{-1-\sqrt{97}}{6}< 0\left(l\right)\end{matrix}\right.\)
\(\Rightarrow\sqrt{x+3}=\frac{\sqrt{97}-1}{6}\Rightarrow x=\left(\frac{\sqrt{97}-1}{6}\right)^2-3=...\)
Đề bài sai, pt này không giải được
