Đáp án:
$\begin{array}{l}
3)a)\left| x \right| + x = 2x\\
\Rightarrow \left| x \right| = 2x - x\\
\Rightarrow \left| x \right| = x\\
\Rightarrow x \ge 0\\
Vậy\,x \ge 0\\
b)\frac{a}{b} = \frac{c}{d} = k\\
\Rightarrow \left\{ \begin{array}{l}
a = b.k\\
c = d.k
\end{array} \right.\\
\left( {a + 2c} \right).\left( {b + d} \right)\\
= \left( {b.k + 2.d.k} \right).\left( {b + d} \right)\\
= k.\left( {b + 2d} \right).\left( {b + d} \right)\\
\left( {a + c} \right).\left( {b + 2d} \right)\\
= \left( {b.k + d.k} \right)\left( {b + 2.d} \right)\\
= k.\left( {b + d} \right).\left( {b + 2d} \right)\\
\Rightarrow \left( {a + 2c} \right).\left( {b + d} \right) = \left( {a + c} \right).\left( {b + 2d} \right)\\
= \left( {k.\left( {b + d} \right).\left( {b + 2d} \right)} \right)
\end{array}$
