\(a,\sqrt{x^2-10x+25}=2\)
\(\Leftrightarrow\sqrt{\left(x-5\right)^2}=2\)
\(\Leftrightarrow\left|x-5\right|=2\)
\(\left|x-5\right|=\left\{{}\begin{matrix}x-5khix\ge5\\-x+5khix< 5\end{matrix}\right.\)
Với \(x\ge5\Rightarrow x-5=2\Rightarrow x=7\left(tmdk\right)\)
Với \(x< 5\Rightarrow-x+5=2\Rightarrow x=3\left(tmdk\right)\)
Vậy \(S=\left\{7;3\right\}\)
\(b,\sqrt{x^2}=3x-2\)
\(\Leftrightarrow\left|x\right|=3x-2\)
\(\left|x\right|=\left\{{}\begin{matrix}xkhix\ge0\\-xkhix< 0\end{matrix}\right.\)
Với \(x\ge0\Rightarrow x=3x-2\Rightarrow x=1\left(tmdk\right)\)
Với \(x< 0\Rightarrow-x=3x-2\Rightarrow x=\dfrac{1}{2}\left(ktmdk\right)\)
Vậy \(S=\left\{1\right\}\)
\(c,\sqrt{4x^2-12x+9}=x+7\)
\(\Leftrightarrow\sqrt{\left(2x-3\right)^2}=x+7\)
\(\Leftrightarrow\left|2x-3\right|=x+7\)
\(\left|2x-3\right|=\left\{{}\begin{matrix}2x-3khix\ge\dfrac{3}{2}\\-2x+3khix>\dfrac{3}{2}\end{matrix}\right.\)
Với \(x\ge\dfrac{2}{3}\Rightarrow2x-3=x+7\Rightarrow x=10\left(tmdk\right)\)
Với \(x>\dfrac{2}{3}\Rightarrow-2x+3=x+7\Rightarrow x=-3x=4\Rightarrow x=-\dfrac{4}{3}\left(ktmdk\right)\)
Vậy \(S=\left\{10\right\}\)
\(a,\sqrt{x^2-10x+25}=2\)
\(\Leftrightarrow\sqrt{\left(x-5\right)^2}=2\)
\(\Leftrightarrow x-5=2\)
\(\Leftrightarrow x=7\)
Vậy \(S=\left\{7\right\}\)
\(b,\sqrt{x^2}=3x-2\)
\(\Leftrightarrow x=3x-2\)
\(\Leftrightarrow x-3x=-2\)
\(\Leftrightarrow-2x=-2\)
\(\Leftrightarrow x=1\)
Vậy \(S=\left\{1\right\}\)
\(c,\sqrt{4x^2-12x+9}=x+7\)
\(\Leftrightarrow\sqrt{\left(2x-3\right)^2}=x+7\)
\(\Leftrightarrow2x-3=x+7\)
\(\Leftrightarrow x=10\)
Vậy \(S=\left\{10\right\}\)
