Question

Giải phương trình :

1. \(x^2-3x-10+3\sqrt{x\left(x-3\right)}=0\)

2. \(\sqrt{\left(\sqrt{x-1}+1\right)^2}+\sqrt{\left(2-\sqrt{x-1}\right)^2}=5\)

Answer 1

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)