Đặt \(sinx+cosx=\sqrt{2}sin\left(x+\frac{\pi}{4}\right)=t\) \(\Rightarrow2sinx.cosx=t^2-1\)
Do \(x\in\left[0;\frac{\pi}{2}\right]\Rightarrow x+\frac{\pi}{4}\in\left[\frac{\pi}{4};\frac{3\pi}{4}\right]\) \(\Rightarrow\frac{\sqrt{2}}{2}\le sin\left(x+\frac{\pi}{4}\right)\le1\)
\(\Rightarrow1\le t\le\sqrt{2}\)
Pt trở thành: \(m\left(t+1\right)=t^2\Leftrightarrow m=\frac{t^2}{t+1}\)
Xét \(f\left(t\right)=\frac{t^2}{t+1}\) trên \(\left[1;\sqrt{2}\right]\)
Có \(f\left(t\right)-\frac{1}{2}=\frac{t^2}{t+1}-\frac{1}{2}=\frac{\left(t-1\right)\left(2t+1\right)}{2\left(t+1\right)}\ge0\Rightarrow f\left(t\right)\ge\frac{1}{2}\)
\(f\left(t\right)-2\sqrt{2}+2=\frac{t^2}{t+1}-2\sqrt{2}+2=\frac{\left(t-\sqrt{2}\right)\left(t+2-\sqrt{2}\right)}{t+1}\le0\Rightarrow f\left(t\right)\le2\sqrt{2}-2\)
\(\Rightarrow\frac{1}{2}\le m\le2\sqrt{2}-2\)
