Ta có : \(4x^2-12x+9=\left(5-x\right)^2\)
<=> \(\left(2x-3\right)^2=\left(5-x\right)^2\)
<=> \(\left(2x-3\right)^2-\left(5-x\right)^2=0\)
<=> \(\left(2x-3-5+x\right)\left(2x-3+5-x\right)=0\)
<=> \(\left(3x-8\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-8=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{8}{3}\\x=-2\end{matrix}\right.\)
\(4x^2-12x+9=\left(5-x\right)^2\Leftrightarrow4x^2-12x+9=25-10x+x^2\)
\(\Leftrightarrow4x^2-12x+9-25+10x-x^2+\Leftrightarrow3x^2-2x-16=0\)
\(\Delta'=\left(-1\right)^2-3.\left(-16\right)=1+48=49>0\)
\(\Rightarrow\) phương trình có 2 nghiệm phân biệt
\(x_1=\dfrac{1+\sqrt{49}}{3}=\dfrac{1+7}{3}=\dfrac{8}{3}\) \(4x^2-12x+9=\left(5-x\right)^2\Leftrightarrow4x^2-12x+9=25-10x+x^2\)
\(x_2=\dfrac{1-\sqrt{49}}{3}=\dfrac{1-7}{3}=\dfrac{-6}{3}=-2\)
vậy \(x=\dfrac{8}{3};x=-2\)
\(4x^2-12x+9=\left(5-x\right)^2\)
\(\Leftrightarrow\left(2x\right)^2-2\cdot2x\cdot3+3^2=\left(5-x\right)^2\)
\(\Leftrightarrow\left(2x-3\right)^2=\left(5-x\right)^2\)
\(\Leftrightarrow\left(2x-3\right)^2-\left(5-x\right)^2=0\)
\(\Leftrightarrow\left(2x-3+5-x\right)\left(2x-3-5+x\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(3x-8\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\Leftrightarrow x=2\\3x-8=0\Leftrightarrow x=\dfrac{8}{3}\end{matrix}\right.\)
