Đáp án:
$\begin{array}{l}
C1)1){\left( {x - 2} \right)^2} = 9\\
\Leftrightarrow \left[ \begin{array}{l}
x - 2 = 3\\
x - 2 = - 3
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 5\\
x = - 1
\end{array} \right.\\
Vậy\,x = - 1;x = 5\\
2)\left\{ \begin{array}{l}
x + 3 = y\\
2x - y = - 2
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x - y = 3\\
2x - y = - 2
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x = - 5\\
y = x + 3 = - 2
\end{array} \right.\\
Vậy\,\left( {x;y} \right) = \left( { - 5; - 2} \right)\\
C2)1)x \ge 0;x\# 4\\
A = \left( {\dfrac{2}{{x - 4}} + \dfrac{1}{{\sqrt x + 2}}} \right):\dfrac{1}{{\sqrt x + 2}} - 1\\
= \dfrac{{2 + \sqrt x - 2}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right)}}.\left( {\sqrt x + 2} \right) - 1\\
= \dfrac{{\sqrt x }}{{\sqrt x - 2}} - 1\\
= \dfrac{{\sqrt x - \sqrt x + 2}}{{\sqrt x - 2}}\\
= \dfrac{2}{{\sqrt x - 2}}\\
2)\left( P \right):y = \dfrac{{{x^2}}}{4}\\
\Leftrightarrow \left\{ \begin{array}{l}
Khi:x = 2 \Leftrightarrow y = \dfrac{{{2^2}}}{4} = 1\\
Khi:x = - 4 \Leftrightarrow y = \dfrac{{{{\left( { - 4} \right)}^2}}}{4} = 4
\end{array} \right.\\
\Leftrightarrow A\left( {2;1} \right);B\left( { - 4;4} \right) \in \left( d \right)\\
Goi:\left( d \right):y = a.x + b\\
\Leftrightarrow \left\{ \begin{array}{l}
1 = 2a + b\\
4 = - 4a + b
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
6a = - 3\\
b = 1 - 2a
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
a = - \dfrac{1}{2}\\
b = 2
\end{array} \right.\\
Vậy\,\left( d \right):y = - \dfrac{1}{2}x + 2
\end{array}$
