\(2\sqrt{x^2+3}-\sqrt{8+2x-x^2}=x
\)
<=> \(2\sqrt{x^2+3}=x+\sqrt{8+2x-x^2}\) ( ĐK : 0 ≤ x ≤ 4)
=> \(\left(2\sqrt{x^2+3}\right)^2=\left(x+\sqrt{8+2x-x^2}\right)^2\)
<=> \(4\left(x^2+3\right)=x^2+8+2x-x^2+2x\sqrt{8+2x-x^2}\)
<=> \(4x^2-2x+4=2x\sqrt{8+2x-x^2}\)
<=> \(2x^2-x+2=x\sqrt{8+2x-x^2}\)
=> \(\left(2x^2-x+2\right)^2=\left(x\sqrt{8+2x-x^2}\right)^2\)
<=> \(4x^4+x^2+4-4x^3-4x+8x^2=x^2\left(8+2x-x^2\right)\)
<=> \(5x^4-6x^3+x^2-4x+4=0\)
<=> \(\left(x-1\right)^2\left(5x^2+4x+4\right)=0\)
=> \(\left[{}\begin{matrix}\left(x-1\right)^2=0\\5x^2+4x+4=0\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x=1\left(tm\right)\\\Delta=-64\left(loại\right)\end{matrix}\right.\)
Vậy x =1
