\(\sqrt{x+2}=3\) (ĐK: \(x\ge-2\))
\(\Leftrightarrow x+2=3^2\)
\(\Leftrightarrow x+2=9\)
\(\Leftrightarrow x=9-2\)
\(\Leftrightarrow x=7\left(tm\right)\)
Vậy: \(x=7\)
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\(\sqrt{4x^2-4x+1}=5\)
\(\Leftrightarrow\sqrt{\left(2x\right)^2-2\cdot2x\cdot1+1}=5\)
\(\Leftrightarrow\sqrt{\left(2x-1\right)^2}=5\)
\(\Leftrightarrow\left|2x-1\right|=5\)
TH1: \(\left|2x-1\right|=2x-1\) với \(2x-1\ge0\Leftrightarrow x\ge\dfrac{1}{2}\)
Pt trở thành:
\(2x-1=5\) (ĐK: \(x\ge\dfrac{1}{2}\))
\(\Leftrightarrow2x=6\)
\(\Leftrightarrow x=3\left(tm\right)\)
TH2: \(\left|2x-1\right|=-\left(2x-1\right)\) với \(2x-1< 0\Leftrightarrow x< \dfrac{1}{2}\)
Pt trở thành:
\(-\left(2x-1\right)=5\) (ĐK: \(x< \dfrac{1}{2}\))
\(\Leftrightarrow-2x+1=5\)
\(\Leftrightarrow-2x=4\)
\(\Leftrightarrow x=-2\left(tm\right)\)
Vậy: \(S=\left\{3;-2\right\}\)
