Đáp án:
C5:
c) \({\left( {x + 5} \right)^2}\)
Giải thích các bước giải:
\(\begin{array}{l}
C5:\\
a)2\left( {{x^2} + 2x} \right) + 1\\
= 2\left( {{x^2} + 2x + 1 - 1} \right) + 1\\
= 2{\left( {x + 1} \right)^2} - 1\\
= \left[ {\sqrt 2 \left( {x + 1} \right) - 1} \right].\left[ {\sqrt 2 \left( {x + 1} \right) + 1} \right]\\
b){x^2} - 6x + 9\\
= {\left( {x - 3} \right)^2}\\
c){x^2} + 10x + 25\\
= {\left( {x + 5} \right)^2}\\
C6:\\
\dfrac{{{n^2}}}{{n + 2}} = \dfrac{{{n^2} - 4 + 4}}{{n + 2}} = \dfrac{{\left( {n - 2} \right)\left( {n + 2} \right) + 4}}{{n + 2}}\\
= n - 2 + \dfrac{4}{{n + 2}}\\
\dfrac{{{n^2}}}{{n + 2}} \in Z\\
\Leftrightarrow \dfrac{4}{{n + 2}} \in Z\\
\Leftrightarrow n + 2 \in U\left( 4 \right)\\
\to \left[ \begin{array}{l}
n + 2 = 4\\
n + 2 = - 4\\
n + 2 = 2\\
n + 2 = - 2\\
n + 2 = 1\\
n + 2 = - 1
\end{array} \right. \to \left[ \begin{array}{l}
n = 2\\
n = - 6\\
n = 0\\
n = - 4\\
n = - 1\\
n = - 3
\end{array} \right.
\end{array}\)
