Question

Giải phương trình:

d)\(\dfrac{3x-1}{18x^2+3x-28}-\dfrac{4x}{24x^2+23x-12}=\dfrac{3}{48x^2-74x+21}\)

Answer 1

\(\Leftrightarrow\dfrac{3x-1}{\left(6x-7\right)\left(3x+4\right)}-\dfrac{4x}{\left(8x-3\right)\left(3x+4\right)}=\dfrac{3}{\left(8x-3\right)\left(6x-7\right)}\)

=>(3x-1)(8x-3)-4x(6x-7)=3(3x+4)

=>24x^2-9x-8x+3-24x^2+28x=9x+12

=>11x+3=9x+12

=>2x=9

=>x=9/2