Đk:\(x\le-3;x\ge1\)
Pt\(\Leftrightarrow\sqrt{4x^2-2x+3}=x+\sqrt{x^2+2x-3}\)
Đặt \(a=\sqrt{x^2+2x-3}\left(a\ge0\right)\)
\(\Leftrightarrow a^2-x^2=2x-3\)\(\Leftrightarrow5x^2-a^2=4x^2-2x+3\)
Pttt:\(\sqrt{5x^2-a^2}=x+a\)\(\Leftrightarrow\left\{{}\begin{matrix}x+a\ge0\\5x^2-a^2=\left(a+x\right)^2\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x+a\ge0\\4x^2-2ax-2a^2=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+a\ge0\\\left[{}\begin{matrix}x=a\\2x=-a\end{matrix}\right.\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x+\sqrt{x^2+2x-3}\ge0\\\left[{}\begin{matrix}x=\sqrt{x^2+2x-3}\\2x=-\sqrt{x^2+2x-3}\end{matrix}\right.\end{matrix}\right.\)
TH1:\(\left\{{}\begin{matrix}x+\sqrt{x^2+2x-3}\ge0\\x=\sqrt{x^2+2x-3}\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x+\sqrt{x^2+2x-3}\ge0\\x\ge0\\x^2=x^2+2x-3\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x+\sqrt{x^2+2x-3}\ge0\\x\ge0\\x=\dfrac{3}{2}\end{matrix}\right.\)
\(\Rightarrow x=\dfrac{3}{2}\)(thỏa mãn đk)
TH2:\(\left\{{}\begin{matrix}x+\sqrt{x^2+2x-3}\ge0\\2x=-\sqrt{x^2+2x-3}\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x+\sqrt{x^2+2x-3}\ge0\\x\le0\\4x^2=x^2+2x-3\end{matrix}\right.\)\(\Rightarrow x\in\varnothing\)
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