\(x^2-4x+4=4x^2\)
\(=>\left(x-2\right)^2=4x^2\)
\(=>x-2=4x\)
\(=>x-2-4x=0\)
\(=>-3x-2=0\)
\(=>3x=-2\)
\(=>x=-\dfrac{2}{3}\)
#Quiz_a bạn trên làm sai rồi :>
a,\(x^2-4x+4=4x^2\)
\(\Leftrightarrow\left(x-2\right)^2-\left(2x\right)^2=0\)
\(\Leftrightarrow\left(-x-2\right)\left(3x-2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(3x-2\right)=0\)
\(\Rightarrow\left\{{}\begin{matrix}x=-2\\x=\dfrac{2}{3}\end{matrix}\right.\)
Vậy ...
b) (2x-1)(x2+x+5) =(1-2x)(4x+1)
=> (2x-1)(x2+x+5) -(1-2x)(4x+1)=0
=> (2x-1)(x2+x+5) +(2x-1)(4x+1) =0
=>(2x-1)(x2+x+5+4x+1) =0
=> (2x-1)(x2 +5x+6) =0
=> (2x-1)(x2 +3x+2x+6) =0
=> (2x-1)(x+3)(x+2) =0
=>\(\left[{}\begin{matrix}2x-1=0\Rightarrow x=\dfrac{1}{2}\\x+3=0\Rightarrow x=-3\\x+2=0\Rightarrow x=-2\end{matrix}\right.\)
Vậy x\(\in\){\(\dfrac{1}{2},-3,-2\)}