Đáp án:
$\begin{array}{l}
a)\left( {2x + 5} \right)\left( {x - 4} \right) = \left( {x - 5} \right)\left( {4 - x} \right)\\
= > \left( {2x + 5} \right)\left( {x - 4} \right) - \left( {x - 5} \right)\left( {4 - x} \right) = 0\\
= > \left( {2x + 5} \right)\left( {x - 4} \right) + \left( {x - 4} \right)\left( {x - 5} \right) = 0\\
= > \left( {x - 4} \right)\left( {2x + 5 + x - 5} \right) = 0\\
= > \left( {x - 4} \right).3x = 0\\
= > x = 4/x = 0\\
Vậy\,x = 4;x = 0\\
b)Dkxd:x\# - 1\\
\dfrac{{2x - 1}}{{{x^3} + 1}} = \dfrac{2}{{{x^2} - x + 1}} - \dfrac{1}{{x + 1}}\\
= > \dfrac{{2x - 1 - 2\left( {x + 1} \right) + \left( {{x^2} - x + 1} \right)}}{{\left( {x + 1} \right)\left( {{x^2} - x + 1} \right)}} = 0\\
= > 2x - 1 - 2x - 2 + {x^2} - x + 1 = 0\\
= > {x^2} - x - 2 = 0\\
= > \left( {x - 2} \right)\left( {x + 1} \right) = 0\\
= > x = 2\left( {tm} \right)/x = - 1\left( {ktm} \right)\\
Vậy\,x = 2\\
c)Dkxd:x\# 0\\
Dat:x + \dfrac{1}{x} = a\\
= > {a^2} + 2a - 8 = 0\\
= > \left( {a - 2} \right)\left( {a + 4} \right) = 0\\
TH1:a - 2 = 0\\
= > x + \dfrac{1}{x} - 2 = 0\\
= > {x^2} - 2x + 1 = 0\\
= > x = 1\left( {tm} \right)\\
TH2:a + 4 = 0\\
= > x + \dfrac{1}{x} + 4 = 0\\
= > {x^2} + 4x + 1 = 0\left( {ktm} \right)\\
Vậy\,x = 1
\end{array}$
