Đáp án:
$\begin{array}{l}
1d)\\
\dfrac{{4x - 3}}{3} + \dfrac{x}{6} = \dfrac{{x - 4}}{2}\\
= > \dfrac{{2\left( {4x - 3} \right) + x}}{6} = \dfrac{{3\left( {x - 4} \right)}}{6}\\
= > 8x - 6 + x = 3x - 12\\
= > 6x = - 6\\
= > x = - 1\\
Vậy\,x = - 1\\
c)Dkxd:x\# 4;x\# - 4\\
\dfrac{{x + 7}}{{x - 4}} - \dfrac{7}{{x + 4}} = \dfrac{{56}}{{{x^2} - 16}}\\
= > \dfrac{{\left( {x + 7} \right)\left( {x + 4} \right) - 7\left( {x - 4} \right)}}{{\left( {x + 4} \right)\left( {x - 4} \right)}} = \dfrac{{56}}{{\left( {x - 4} \right)\left( {x + 4} \right)}}\\
= > {x^2} + 11x + 28 - 7x + 28 = 56\\
= > {x^2} + 4x = 0\\
= > x\left( {x + 4} \right) = 0\\
= > x = 0\left( {do:x\# - 4} \right)\\
Vậy\,x = 0\\
e)\backslash 2x - 5\backslash = x + 1\\
+ TH1:2x - 5 = x + 1\\
= > x = 6\\
+ TH2:2x - 5 = - x - 1\\
= > 3x = 4\\
= > x = \dfrac{4}{3}\\
Vậy\,x = 6;x = \dfrac{4}{3}
\end{array}$
