\(\left(4x+1\right)\left(x-3\right)-\left(x-7\right)\left(4x-1\right)=15\\ \Leftrightarrow4x^2+x-12x-3-\left(4x^2-28x-x+7\right)-15=0\\ \Leftrightarrow4x^2-11x-3-4x^2+29x-7-15=0\\ \Leftrightarrow18x=25\\ \Leftrightarrow x=\dfrac{25}{18}\)
Vậy \(x=\dfrac{25}{18}\)
Lời giải:
$(4x+1)(x-3)-(x-7)(4x-1)=15$
$\Leftrightarrow (4x^2-11x-3)-(4x^2-29x+7)=15$
$\Leftrightarrow 18x-10=15$
$\Leftrightarrow 18x=25$
$\Leftrightarrow x=\frac{25}{18}$