Đáp án:
B1:
\(\left[ \begin{array}{l}
x = 1\\
x = 0
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
B1:\\
2x + 4 \vdots x + 1\\
\to 2\left( {x + 1} \right) + 2 \vdots x + 1\\
\to 2 \vdots x + 1\\
\to x + 1 \in U\left( 2 \right)\\
\to \left[ \begin{array}{l}
x + 1 = 2\\
x + 1 = 1
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 1\\
x = 0
\end{array} \right.\\
B2:\\
a)\dfrac{{ - 6}}{{120}} = - \dfrac{1}{{20}} = - \dfrac{5}{{20.5}} = - \dfrac{5}{{100}}\\
\dfrac{4}{{25}} = \dfrac{{4.4}}{{25.4}} = \dfrac{{16}}{{100}}\\
b) - \dfrac{3}{{18}} = - \dfrac{1}{6} = - \dfrac{{3.5}}{{6.15}} = - \dfrac{{15}}{{90}}\\
- \dfrac{{135}}{{450}} = - \dfrac{3}{{10}} = - \dfrac{{3.9}}{{90}} = - \dfrac{{27}}{{90}}\\
- \dfrac{{11}}{{45}} = - \dfrac{{11.2}}{{90}} = - \dfrac{{22}}{{90}}\\
c) - \dfrac{8}{{11}} = - \dfrac{{8.2.6}}{{11.12}} = - \dfrac{{96}}{{132}}\\
- \dfrac{1}{{12}} = - \dfrac{{11}}{{132}}\\
\dfrac{{35}}{{66}} = \dfrac{{35.2}}{{132}} = \dfrac{{70}}{{132}}
\end{array}\)
