Đáp án:
$\begin{array}{l}
d:y = a.x + b\\
a)A\left( {\sqrt 2 ;1} \right);B\left( {0;1 + 3\sqrt 2 } \right)\\
\Leftrightarrow \left\{ \begin{array}{l}
1 = \sqrt 2 a + b\\
b = 1 + 3\sqrt 2
\end{array} \right.\\
\Leftrightarrow 1 = \sqrt 2 a + 1 + 3\sqrt 2 \\
\Leftrightarrow \sqrt 2 .a = - 3\sqrt 2 \\
\Leftrightarrow a = - 3\\
Vậy\,a = - 3\\
b)Xet:2x - 1 = - x + 2\\
\Leftrightarrow 2x + x = 2 + 1\\
\Leftrightarrow 3x = 3\\
\Leftrightarrow x = 1\\
\Leftrightarrow y = - x + 2 = 1\\
\Leftrightarrow {d_1} \cap {d_2} = \left( {1;1} \right)\\
\Leftrightarrow \left( {1;1} \right);C\left( {\dfrac{1}{2};\dfrac{{ - 1}}{4}} \right) \in d\\
\Leftrightarrow \left\{ \begin{array}{l}
1 = a + b\\
- \dfrac{1}{4} = \dfrac{1}{2}a + b
\end{array} \right.\\
\Leftrightarrow a - \dfrac{1}{2}a = 1 - \left( {\dfrac{{ - 1}}{4}} \right)\\
\Leftrightarrow \dfrac{1}{2}a = \dfrac{5}{4}\\
\Leftrightarrow a = \dfrac{5}{2}\\
Vậy\,a = \dfrac{5}{2}\\
c){d_3}:y = \left( {2m + 1} \right).x + 2m - 1\forall m\\
\Leftrightarrow 2mx + x + 2m - 1 = y\forall m\\
\Leftrightarrow \left( {2x + 2} \right).m = y - x + 1\forall m\\
\Leftrightarrow \left\{ \begin{array}{l}
2x + 2 = 0\\
y - x + 1 = 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x = - 1\\
y = x - 1 = - 2
\end{array} \right.\\
\Leftrightarrow \left( { - 1; - 2} \right);D\left( {0; - 1} \right) \in d\\
\Leftrightarrow \left\{ \begin{array}{l}
- 2 = a.\left( { - 1} \right) + b\\
b = - 1
\end{array} \right.\\
\Leftrightarrow - 2 = - a - 1\\
\Leftrightarrow a = 2 - 1 = 1\\
Vậy\,a = 1
\end{array}$
