a, \(\sqrt{\left(x-3\right)^2}=3-x\Leftrightarrow\left|x-3\right|=3-x\)ĐK : \(x\le3\)
TH1 : \(x-3=3-x\Leftrightarrow x=3\)
TH2 : \(x-3=x-3\Leftrightarrow0x=0\)
Vậy pt có vô số nghiệm
b, \(\sqrt{x+2\sqrt{x-1}}=2\Leftrightarrow\sqrt{x-1+2\sqrt{x-1}+1}=2\)ĐK : \(x\ge1\)
\(\Leftrightarrow\sqrt{x-1}+1=2\Leftrightarrow x-1=1\Leftrightarrow x=2\)( tm )
a,\(\sqrt{\left(x-3\right)^2}=3-x\)(Đk:x≤3)
\(\Leftrightarrow\left|x-3\right|=3-x\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=3-x\\x-3=x-3\end{matrix}\right.\)
\(\Rightarrow2x=6\)
\(\Leftrightarrow x=3\)(N)
Vậy ....
b,\(\sqrt{x+2\sqrt{x-1}}=2\)(Đk x≥1)
\(\Leftrightarrow\sqrt{\left(\sqrt{x-1}+1\right)^2}=2\)
⇔\(\left|\sqrt{x-1}+1\right|=2\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-1}+1=2\\\sqrt{x-1}+1=-2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-1}=1\\\sqrt{x-1}=-3\left(VL\right)\end{matrix}\right.\)
\(\Leftrightarrow x-1=1\)
\(\Leftrightarrow x=2\)(n)
Vậy....
