Question

Giải phương trình sau:

a)x² -4x +4 = 4x²

b)(2x-1).(x²+x+5)=(1-2x).(4x+1)

Answer 1

\(x^2-4x+4=4x^2\)

\(=>\left(x-2\right)^2=4x^2\)

\(=>x-2=4x\)

\(=>x-2-4x=0\)

\(=>-3x-2=0\)

\(=>3x=-2\)

\(=>x=-\dfrac{2}{3}\)

Answer 2

#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 ...

Answer 3

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\(\in\){\(\dfrac{1}{2},-3,-2\)}