Giải thích các bước giải:
\(\begin{array}{l}
1)a)DK:x \ne \pm 2\\
\dfrac{x}{{{x^2} - 4}} + \dfrac{{3\left( {x - 2} \right)}}{{{x^2} - 4}} = \dfrac{{{x^2} - 4}}{{{x^2} - 4}}\\
\Leftrightarrow x + 3x - 6 = {x^2} - 4\\
\Leftrightarrow {x^2} - 4x - 2 = 0\\
\Leftrightarrow x = 2 \pm \sqrt 6 \left( {tm} \right)\\
b)\sqrt {x + 8} = x + 2\\
\Leftrightarrow \left\{ \begin{array}{l}
x + 2 \ge 0\\
x + 8 = {\left( {x + 2} \right)^2}
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x \ge - 2\\
{x^2} + 3x - 4 = 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x \ge - 2\\
\left[ \begin{array}{l}
x = 1\\
x = - 4
\end{array} \right.
\end{array} \right. \Rightarrow x = 1\\
c)\left\{ \begin{array}{l}
x = y - 1\\
2\left( {y - 1} \right) + 3y = 8
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x = y - 1\\
5y = 10
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
y = 2\\
x = 1
\end{array} \right.\\
d)\left\{ \begin{array}{l}
3x - y = 2\\
x - 5y = - 4\\
x + 2y - z = 2
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x = 1\\
y = 1\\
z = 1
\end{array} \right.\\
3)\overrightarrow {MN} = \overrightarrow {MA} + \overrightarrow {AC} + \overrightarrow {CN} \\
\overrightarrow {MN} = \overrightarrow {MB} + \overrightarrow {BD} + \overrightarrow {DN} \\
\Rightarrow 2\overrightarrow {MN} = \left( {\overrightarrow {MA} + \overrightarrow {MB} } \right) + \overrightarrow {AC} + \overrightarrow {BD} + \left( {\overrightarrow {DN} + \overrightarrow {CN} } \right)\\
\Rightarrow 2\overrightarrow {MN} = \overrightarrow {AC} + \overrightarrow {BD}
\end{array}\)
Em tách hỏi từng câu thôi nhé
