Question

giải pt : \(2x^2-5x-7+\left(x-1\right)\sqrt{x+1}=0\)

Answer 1

ĐKXĐ: \(x\ge-1\)

\(\left(x+1\right)\left(2x-7\right)+\left(x-1\right)\sqrt{x+1}=0\)

\(\Leftrightarrow\sqrt{x+1}\left(\left(2x-7\right)\sqrt{x+1}+x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\Rightarrow x=-1\\\left(2x-7\right)\sqrt{x+1}+x-1=0\left(1\right)\end{matrix}\right.\)

Xét (1), đặt \(\sqrt{x+1}=t\ge0\Rightarrow x=t^2-1\)

\(\left(1\right)\Leftrightarrow\left(2t^2-9\right)t+t^2-2=0\)

\(\Leftrightarrow2t^3+t^2-9t-2=0\)

\(\Leftrightarrow\left(t-2\right)\left(2t^2+5t+1\right)=0\)

\(\Rightarrow t-2=0\) (do \(2t^2+5t+1>0\) \(\forall t\ge0\))

\(\Rightarrow t=2\Rightarrow\sqrt{x+1}=2\Rightarrow x=3\)

Vậy nghiệm của pt là \(x=\left\{-1;3\right\}\)