ĐK: \(\forall x\in R\)
PT\(\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{5}{2}\\x^2-6x+9=4x^2-20x+25\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{5}{2}\\3x^2-14x+16=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\left(l\right)\\x=\dfrac{8}{3}\left(tm\right)\end{matrix}\right.\)
Điều kiện :
\(\left\{{}\begin{matrix}x^2-6x+9\ge0\\2x-5\ge0\end{matrix}\right.\)⇔ \(x\ge\dfrac{5}{2}\)
Ta có :
\(\left(\sqrt{x^2-6x+9}\right)^2=\left(2x-5\right)^2\)
⇔ \(x^2-6x+9=4x^2-20x+25\)
⇔ \(3x^2-14x+16=0\)
⇔\(\left\{{}\begin{matrix}x=2\left(loại\right)\\x=\dfrac{8}{3}\left(tm\right)\end{matrix}\right.\)
Vì \(VT\ge0\Rightarrow VP\ge0\Rightarrow x\ge\dfrac{5}{2}\)
\(\sqrt{x^2-6x+9}=2x-5\Rightarrow\sqrt{\left(x-3\right)^2}=2x-5\)
\(\Rightarrow\left|x-3\right|=2x-5\)
Xét \(x\ge3\Rightarrow x-3=2x-5\Rightarrow x-2=0\Rightarrow x=2\) (loại)
Xét \(x< 3\Rightarrow\dfrac{5}{2}\le x< 3\Rightarrow3-x=2x-5\Rightarrow3x-8=0\Rightarrow x=\dfrac{8}{3}\)
Vậy \(x=\dfrac{8}{3}\) là nghiệm của pt...
Lời giải:
PT \(\Rightarrow \left\{\begin{matrix} 2x-5\geq 0\\ x^2-6x+9=(2x-5)^2\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\geq \frac{5}{2}\\ 3x^2-14x+16=0\end{matrix}\right.\)
\(\Leftrightarrow \left\{\begin{matrix} x\geq \frac{5}{2}\\ (3x-8)(x-2)=0\end{matrix}\right.\Rightarrow x=\frac{8}{3}\)
Ta có: \(\sqrt{x^2-6x+9}=2x-5\)
\(\Leftrightarrow\left|x-3\right|=2x-5\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=2x-5\left(x\ge3\right)\\x-3=5-2x\left(x< 3\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x-2x=-5+3\\x+2x=5+3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}-x=-2\\3x=8\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\left(loại\right)\\x=\dfrac{8}{3}\left(nhận\right)\end{matrix}\right.\)
Vậy: \(S=\left\{\dfrac{8}{3}\right\}\)
