1) \(\sqrt{x^2-2x+1}=2\)
\(\Leftrightarrow\sqrt{x^2-2\cdot x\cdot1+1^2}=2\)
\(\Leftrightarrow\sqrt{\left(x-1\right)^2}=2\)
\(\Leftrightarrow\left|x-1\right|=2\)
TH1: \(x-1=x-1\) với \(x\ge1\)
Pt trở thành:
\(x-1=2\) (ĐK: \(x\ge1\))
\(\Leftrightarrow x=2+1\)
\(\Leftrightarrow x=3\left(tm\right)\)
TH2: \(x-1=-\left(x-1\right)\) với \(x< 1\)
Pt trở thành:
\(-\left(x-1\right)=2\) (ĐK: \(x< 1\))
\(\Leftrightarrow x-1=-2\)
\(\Leftrightarrow x=-2+1\)
\(\Leftrightarrow x=-1\left(tm\right)\)
Vậy: \(S=\left\{-1;3\right\}\)
2) \(\dfrac{1}{2}\sqrt{16x-32}-\dfrac{1}{3}\sqrt{9x-18}+\sqrt{25x-50}=6\) (ĐK: \(x\ge2\))
\(\Leftrightarrow\dfrac{1}{2}\sqrt{16\left(x-2\right)}-\dfrac{1}{3}\sqrt{9\left(x-2\right)}+5\sqrt{x-2}=6\)
\(\Leftrightarrow\dfrac{1}{2}\cdot4\sqrt{x-2}-\dfrac{1}{3}\cdot3\sqrt{x-2}+5\sqrt{x-2}=6\)
\(\Leftrightarrow2\sqrt{x-2}-\sqrt{x-2}+5\sqrt{x-2}=6\)
\(\Leftrightarrow6\sqrt{x-2}=6\)
\(\Leftrightarrow\sqrt{x-2}=\dfrac{6}{6}\)
\(\Leftrightarrow\sqrt{x-2}=1\)
\(\Leftrightarrow x-2=1\)
\(\Leftrightarrow x=1+2\)
\(\Leftrightarrow x=3\left(tmdk\right)\)
Vậy \(x=3\)
1:=>\(\sqrt{\left(x-1\right)^2}=2\)
=>|x-1|=2
=>x-1=2 hoặc x-1=-2
=>x=3 hoặc x=-1
2: \(\Leftrightarrow4\cdot\dfrac{1}{2}\sqrt{x-2}-\dfrac{1}{3}\cdot3\sqrt{x-2}+5\sqrt{x-2}=6\)
=>\(6\sqrt{x-2}=6\)
=>căn x-2=1
=>x-2=1
=>x=3
