a. ĐKXĐ:...
Đặt \(\sqrt{x^2-3x}=t\ge0\)
\(\Rightarrow t^2+3t-10=0\Rightarrow\left[{}\begin{matrix}t=2\\t=-5\left(l\right)\end{matrix}\right.\)
\(\Rightarrow\sqrt{x^2-3x}=2\)
\(\Leftrightarrow x^2-3x=4\)
\(\Rightarrow\left[{}\begin{matrix}x=-1\\x=4\end{matrix}\right.\)
2. ĐKXĐ: ...
\(\Leftrightarrow\left|\sqrt{x-1}+1\right|+\left|2-\sqrt{x-1}\right|=5\)
\(\Leftrightarrow\sqrt{x-1}+\left|2-\sqrt{x-1}\right|=4\)
TH1: \(x\ge5\Rightarrow2-\sqrt{x-1}\le0\)
Pt trở thành: \(\sqrt{x-1}+\sqrt{x-1}-2=4\)
\(\Leftrightarrow\sqrt{x-1}=3\Rightarrow x=10\)
Th2: \(1\le x< 5\Rightarrow2-\sqrt{x-1}>0\)
Pt trở thành: \(\sqrt{x-1}+2-\sqrt{x-1}=4\Leftrightarrow2=4\) (vô nghiệm)
