Question

Tìm x 

a/ x² . ( 2x - 3 ) - 12 + 8x = 0

b/ 2010x² - x = 2011

Answer 1

a: \(x^2\left(2x-3\right)+8x-12=0\)

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

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

mà \(x^2+4>=4>0\forall x\)

nên 2x-3=0

=>2x=3

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

b: \(2010x^2-x=2011\)

=>\(2010x^2-x-2011=0\)

=>\(2010x^2-2011x+2010x-2011=0\)

=>\(\left(2010x-2011\right)\left(x+1\right)=0\)

=>\(\left[{}\begin{matrix}2010x-2011=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2011}{2010}\\x=-1\end{matrix}\right.\)