Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
a,\\
{\left( {x + 2} \right)^2} + {\left( {x - 3} \right)^2} = 2x\left( {x + 7} \right)\\
\Leftrightarrow \left( {{x^2} + 4x + 4} \right) + \left( {{x^2} - 6x + 9} \right) = 2{x^2} + 14x\\
\Leftrightarrow 2{x^2} - 2x + 13 = 2{x^2} + 14x\\
\Leftrightarrow 16x = 13\\
\Leftrightarrow x = \dfrac{{13}}{{16}}\\
b,\\
\left( {x + 3} \right)\left( {{x^2} - 3x + 9} \right) = x\left( {{x^2} + 4} \right) - 1\\
\Leftrightarrow {x^3} + {3^3} = {x^3} + 4x - 1\\
\Leftrightarrow {x^3} + 27 = {x^3} + 4x - 1\\
\Leftrightarrow 4x = 28\\
\Leftrightarrow x = 7\\
c,\\
{\left( {x + 1} \right)^3} + {\left( {x - 1} \right)^3} = 2{x^3}\\
\Leftrightarrow \left( {{x^3} + 3{x^2} + 3x + 1} \right) + \left( {{x^3} - 3{x^2} + 3x - 1} \right) = 2{x^3}\\
\Leftrightarrow 2{x^3} + 6x = 2{x^3}\\
\Leftrightarrow 6x = 0\\
\Leftrightarrow x = 0\\
d,\\
{\left( {2x + 1} \right)^2} = 16\\
\Leftrightarrow \left[ \begin{array}{l}
2x + 1 = 4\\
2x + 1 = - 4
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{3}{2}\\
x = - \dfrac{5}{2}
\end{array} \right.\\
e,\\
{x^2} - 10x = 11\\
\Leftrightarrow {x^2} - 10x - 11 = 0\\
\Leftrightarrow \left( {{x^2} - 11x} \right) + \left( {x - 11} \right) = 0\\
\Leftrightarrow x\left( {x - 11} \right) + \left( {x - 11} \right) = 0\\
\Leftrightarrow \left( {x - 11} \right)\left( {x + 1} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x - 11 = 0\\
x + 1 = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 11\\
x = - 1
\end{array} \right.
\end{array}\)
