Question

(x-2)(3-4x) + (x^2 - 4x +4) = 0

Answer 1

\(\left(x-2\right)\left(3-4x\right)+\left(x^2-4x+4\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(3-4x\right)+\left(x-2\right)^2=0\)

\(\Leftrightarrow\left(x-2\right)\left(3-4x+x-2\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(1-3x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{1}{3}\end{matrix}\right.\)

Vậy..........

Answer 2

\(\left(x-2\right)\left(3-4x\right)+\left(x^2-4x+4\right)=0\)

<=>\(\left(x-2\right)\left(3-4x\right)+\left(x-2\right)^2=0\)

<=>\(\left(x-2\right)\left(3-4x+x-2\right)=0\)

<=>\(\left(x-2\right)\left(-3x+1\right)=0< =>\left[{}\begin{matrix}x-2=0\\-3x+1=0\end{matrix}\right.< =>\left[{}\begin{matrix}x=2\\x=\dfrac{1}{3}\end{matrix}\right.\)

vậy.........