Đáp án:
$\begin{array}{l}
a)\left( {{x^2} + x + 1} \right)\left( {x - 1} \right) = 2{x^3}\\
\Leftrightarrow {x^3} - {1^3} = 2{x^3}\\
\Leftrightarrow {x^3} = - {1^3}\\
\Leftrightarrow x = - 1\\
\text{Vậy}\,x = - 1\\
b){\left( {x + 1} \right)^2} - 2x = 0\\
\Leftrightarrow {x^2} + 2x + 1 - 2x = 0\\
\Leftrightarrow {x^2} + 1 = 0\\
\Leftrightarrow {x^2} = - 1\left( {vn} \right)\\
\text{Vậy}\,pt\,\text{vô}\,\text{nghiệm}\\
c)x\left( {{x^2} + x + 1} \right) - {x^2}\left( {x + 1} \right) - x + 5 = 2\\
\Leftrightarrow {x^3} + {x^2} + x - {x^3} - {x^2} - x + 5 - 2 = 0\\
\Leftrightarrow 3 = 0\left( {vn} \right)\\
\text{Vậy}\,pt\,\text{vô}\,\text{nghiệm}\\
d){\left( {x - 2} \right)^2} - 4 = 0\\
\Leftrightarrow {\left( {x - 2} \right)^2} = {2^2}\\
\Rightarrow \left[ \begin{array}{l}
x - 2 = 2\\
x - 2 = - 2
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x = 4\\
x = 0
\end{array} \right.\\
\text{Vậy}\,x = 4hoac\,x = 0\\
e){x^2} - 2x + 1 = 0\\
\Leftrightarrow {\left( {x - 1} \right)^2} = 0\\
\Leftrightarrow x - 1 = 0\\
\Leftrightarrow x = 1\\
\text{Vậy}\,x = 1\\
f){\left( {2x + 1} \right)^2} - 2\left( {2{x^2} + 3} \right) = 0\\
\Leftrightarrow 4{x^2} + 4x + 1 - 4{x^2} - 6 = 0\\
\Leftrightarrow 4x - 5 = 0\\
\Leftrightarrow x = \dfrac{5}{4}\\
\text{Vậy}\,x = \dfrac{5}{4}\\
a)\left( {3x - 5} \right)\left( {5 - 3x} \right) + 9{\left( {x + 1} \right)^2} = 30\\
\Leftrightarrow - {\left( {3x - 5} \right)^2} + 9\left( {{x^2} + 2x + 1} \right) = 30\\
\Leftrightarrow - \left( {9{x^2} - 30x + 25} \right) + 9{x^2} + 18x + 9 = 30\\
\Leftrightarrow - 9{x^2} + 30x - 25 + 9{x^2} + 18x + 9 = 30\\
\Leftrightarrow 48x = 46\\
\Leftrightarrow x = \dfrac{{23}}{{24}}\\
\text{Vậy}\,x = \dfrac{{23}}{{24}}\\
b){\left( {x + 4} \right)^2} - \left( {x + 1} \right)\left( {x - 1} \right) = 16\\
\Leftrightarrow {x^2} + 8x + 16 - {x^2} + 1 = 16\\
\Leftrightarrow 8x = - 1\\
\Leftrightarrow x = - \dfrac{1}{8}\\
\text{Vậy}\,x = - \dfrac{1}{8}
\end{array}$
