Đáp án:
b) x=-2
Giải thích các bước giải:
\(\begin{array}{l}
a)\left( {x - 2} \right)\left( {{x^2} + 2x + 4} \right) + 2\left( {x - 2} \right)\left( {x + 2} \right) - 5\left( {x - 2} \right) = 0\\
\to \left( {x - 2} \right)\left( {{x^2} + 2x + 4 + 2x + 4 - 5} \right) = 0\\
\to \left[ \begin{array}{l}
x - 2 = 0\\
{x^2} + 4x + 3 = 0
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 2\\
{x^2} + x + 3x + 3 = 0
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 2\\
x\left( {x + 1} \right) + 3\left( {x + 1} \right) = 0
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 2\\
\left( {x + 1} \right)\left( {x + 3} \right) = 0
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 2\\
x = - 1\\
x = - 3
\end{array} \right.\\
b){\left( {x + 2} \right)^2} - \left( {x - 2} \right)\left( {x + 2} \right) = 0\\
\to \left( {x + 2} \right)\left( {x + 2 - x + 2} \right) = 0\\
\to x + 2 = 0\\
\to x = - 2
\end{array}\)
