`~rai~`
\(2(\sin x-\cos x)+1=\sin x\cos x\\\Leftrightarrow 2(\sin x-\cos x)-\sin x\cos x+1=0\quad(1)\\\text{Đặt }\sin x-\cos x=t\quad(|t|\le \sqrt{2})\\\Leftrightarrow t^2=(\sin x-\cos x)^2\\\Leftrightarrow t^2=\sin^2x-2\sin x\cos x\\\Leftrightarrow t^2=\sin^2-2\sin x\cos x+\cos^2x\\\Leftrightarrow t^2=1-2\sin x\cos x\\\Leftrightarrow \sin x\cos x=1-t^2.\\\text{Thay vào (1) được:}\\2t-\dfrac{1-t^2}{2}+1=0\\\Leftrightarrow t^2+4t+1=0\\\Leftrightarrow \left[\begin{array}{I}t=-2+\sqrt{3}\text{(nhận)}\\t=-2-\sqrt{3}\text{(loại)}\end{array}\right.\\\Leftrightarrow \sin x-\cos x=-2+\sqrt{3}\\\to\text{Nghiệm xấu.}\\\to\text{Đáp án sai.}\)
