Đáp án:
$\begin{array}{l}
a)2x + x\left( {x + 1} \right)\left( {x - 1} \right)\\
= \left( {x + 1} \right)\left( {{x^2} - x + 1} \right)\\
\Rightarrow 2x + x\left( {{x^2} - 1} \right) = {x^3} + 1\\
\Rightarrow 2x + {x^3} - x = {x^3} + 1\\
\Rightarrow x = 1\\
Vậy\,x = 1\\
b)\dfrac{{x - 1}}{2} + \dfrac{{x - 1}}{3} - \dfrac{{x - 1}}{6} = 2\\
\Rightarrow \left( {x - 1} \right).\left( {\dfrac{1}{2} + \dfrac{1}{3} + \dfrac{1}{6}} \right) = 2\\
\Rightarrow \left( {x - 1} \right).1 = 2\\
\Rightarrow x - 1 = 2\\
\Rightarrow x = 3\\
Vậy\,x = 3\\
c)\dfrac{{2 + x}}{5} - 0,5.x = \dfrac{{1 - 2x}}{4} + 0,25\\
\Rightarrow \dfrac{{2 + x}}{5} - \dfrac{x}{2} = \dfrac{{1 - 2x}}{4} + \dfrac{1}{4}\\
\Rightarrow \dfrac{{2\left( {2 + x} \right) - 5x}}{{10}} = \dfrac{{ - 2x}}{4}\\
\Rightarrow \dfrac{{4 + 2x - 5x}}{{10}} = - \dfrac{x}{2}\\
\Rightarrow 4 - 3x = - 5x\\
\Rightarrow 2x = - 4\\
\Rightarrow x = - 2\\
Vậy\,x = - 2\\
d)\dfrac{{\left( {2x - 1} \right)\left( {x + 2} \right)}}{3} - \dfrac{{2{x^2} + 1}}{2} = \dfrac{{11}}{2}\\
\Rightarrow \dfrac{{2\left( {2{x^2} + 3x - 2} \right) - 3\left( {2{x^2} + 1} \right)}}{6} = \dfrac{{33}}{6}\\
\Rightarrow 4{x^2} + 6x - 4 - 6{x^2} - 3 - 33 = 0\\
\Rightarrow - 2{x^2} + 6x - 40 = 0\\
\Rightarrow {x^2} - 3x + 20 = 0\left( {vn} \right)
\end{array}$
Vậy pt vô nghiệm
