Question

\(x^2-8x-3+6\sqrt{2x+3}=0\)

Answer 1

ĐKXĐ: \(x\ge-\dfrac{3}{2}\)

\(\left(x^2-6x+9\right)-\left(2x+3-6\sqrt{2x+3}+9\right)=0\)

\(\Leftrightarrow\left(x-3\right)^2-\left(\sqrt{2x+3}-3\right)^2=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-3=\sqrt{2x+3}-3\\x-3=3-\sqrt{2x+3}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{2x+3}\left(x\ge0\right)\\\sqrt{2x+3}=6-x\left(x\le6\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2-2x-3=0\left(x\ge0\right)\\2x+3=x^2-12x+36\left(x\le6\right)\end{matrix}\right.\)

\(\Rightarrow x=...\)