Question

Giải phương trình

\(\sqrt{x+4}-\sqrt{1-x}=\sqrt{1-2x}\)

Answer 1

ĐK: \(-4\le x\le\dfrac{1}{2}\)

\(\sqrt{x+4}=\sqrt{1-x}+\sqrt{1-2x}\)

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

\(\Leftrightarrow2x+1=\sqrt{1-3x+2x^2}\)

\(\Leftrightarrow\left\{{}\begin{matrix}2x+1\ge0\\\left(2x+1\right)^2=2x^2-3x+1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{-1}{2}\\2x^2+7x=0\end{matrix}\right.\)

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

Vậy pt có nghiệm duy nhất \(x=0\)