Question

giải phương trình \(4\sqrt{x+1}=x^2-5x+14\)

Answer 1

ĐKXĐ: \(x\ge-1\)

pt \(\Leftrightarrow-x+4\sqrt{x+1}-5=x^2-5x+14-5\)

\(\Leftrightarrow-\left(x+1-4\sqrt{x+1}+4\right)=x^2-6x+9\)

\(\Leftrightarrow-\left(\sqrt{x+1}-2\right)^2=\left(x-3\right)^2\)

Ta thấy \(VT\le0\forall x\ge-1;VP\ge0\forall x\)

nên pt \(\Leftrightarrow\left\{{}\begin{matrix}\left(\sqrt{x+1}-2\right)^2=0\\\left(x-3\right)^2=0\end{matrix}\right.\)

\(\Leftrightarrow x=3\left(TM\right)\)

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