Question

Tìm x, bt :

a,x{2x-1}+1/3-2/3x=0

b, x^2-4x+{x-4}^2=0

Answer 1

a: \(\Leftrightarrow2x^2-x+\dfrac{1}{3}-\dfrac{2}{3}x=0\)

\(\Leftrightarrow2x^2-\dfrac{5}{3}x+\dfrac{1}{3}=0\)

\(\Leftrightarrow6x^2-5x+1=0\)

\(\Leftrightarrow6x^2-2x-3x+1=0\)

=>(2x-1)(3x-1)=0

=>x=1/3 hoặc x=1/2

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

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

=>x=4 hoặc x=2