Question

4x^{2 -} 12x + 5 = 0

Answer 1

\(4x^2-12x+5=0\Leftrightarrow4x^2-10x-2x+5=0\Leftrightarrow2x\left(2x-5\right)-\left(2x-5\right)=0\)

\(\Leftrightarrow\left(2x-5\right)\left(2x-1\right)=0\)\(\Leftrightarrow\left[{}\begin{matrix}2x-5=0\\2x-1=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=\dfrac{1}{2}\end{matrix}\right.\)

Vậy ...

Answer 2

4x² - 12x + 5 = 0 <=> 4x² - 2x - 10x + 5 =0

<=> 2x ( 2x - 1) - 5 (2x - 1) = 0

<=> (2x-5)(2x-1) = 0

=> \(\left\{{}\begin{matrix}2x-5=0< =>x=2,5\\2x-1=0< =>x=0,5\end{matrix}\right.\)

Vậy với x = 0,5 hoặc x = 2,5 thì ta đc PT trên.