a, ĐKXĐ : \(\left\{{}\begin{matrix}x+2-3\sqrt{2x-5}\ge0\\2x-5\ge0\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}x+2\ge3\sqrt{2x-5}\\2x-5\ge0\end{matrix}\right.\)
=> \(x\ge\frac{5}{2}\)
Ta có : \(Q=\frac{\sqrt{x+2-3\sqrt{2x-5}}}{\sqrt{2}}\)
=> \(Q=\sqrt{\frac{x+2-3\sqrt{2x-5}}{2}}\)
=> \(Q=\sqrt{\frac{2x+4-6\sqrt{2x-5}}{4}}\)
=> \(Q=\sqrt{\frac{2x-5-2.3\sqrt{2x-5}+9}{4}}\)
=> \(Q=\sqrt{\frac{\left(\sqrt{2x-5}-3\right)^2}{4}}\)
=> \(Q=\frac{\left|\sqrt{2x-5}-3\right|}{2}\)
b, - Thay x = \(4,5-\sqrt{3}=\frac{9}{2}-\sqrt{3}\) ta được :
\(Q=\frac{\left|\sqrt{2\left(\frac{9}{2}-\sqrt{3}\right)-5}-3\right|}{2}\) ( còn lại bấm máy nốt nha bạn )
