Question

Tìm x 

\(\sqrt{2x+4}=\dfrac{6x-4}{\sqrt{x^2+4}}+2\sqrt{2-x}\)

Answer 1

ĐKXĐ: \(-2\le x\le2\)

\(\Leftrightarrow\sqrt{2x+4}-2\sqrt{2-x}=\dfrac{6x-4}{\sqrt{x^2+4}}\)

\(\Leftrightarrow\dfrac{6x-4}{\sqrt{2x+4}+2\sqrt{2-x}}=\dfrac{6x-4}{\sqrt{x^2+4}}\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\\sqrt{2x+4}+2\sqrt{2-x}=\sqrt{x^2+4}\left(1\right)\end{matrix}\right.\)

\(\left(1\right)\Leftrightarrow2x+4+8-4x+4\sqrt{2\left(4-x^2\right)}=x^2+4\)

\(\Leftrightarrow4\sqrt{2\left(4-x^2\right)}=x^2+2x-8\)

\(\Leftrightarrow4\sqrt{2\left(4-x^2\right)}=\left(x+4\right)\left(x-2\right)\)

Do \(x\le2\Rightarrow\left\{{}\begin{matrix}VT\ge0\\VP\le0\end{matrix}\right.\)

Dấu "=" xảy ra khi và chỉ khi \(x=2\)

Vậy pt có 2 nghiệm \(x=\left\{\dfrac{2}{3};2\right\}\)