Đáp án:

\(\begin{array}{l}
B5:\\
a)\left\{ \begin{array}{l}
y = 4\\
x = 7
\end{array} \right.\\
b)m \ne  \pm \dfrac{1}{{\sqrt 2 }}\\
B6:\\
a)\left\{ \begin{array}{l}
x = 2\\
y = 0
\end{array} \right.\\
b)\left[ \begin{array}{l}
m = \dfrac{{ - 2 + \sqrt {10} }}{2}\\
m = \dfrac{{ - 2 - \sqrt {10} }}{2}
\end{array} \right.
\end{array}\)

Giải thích các bước giải:

\(\begin{array}{l}
B5:\\
\left\{ \begin{array}{l}
mx - y = 3\\
 - x + 2my = 1
\end{array} \right.\\
 \to \left\{ \begin{array}{l}
mx - y = 3\\
 - mx + 2{m^2}y = m
\end{array} \right.\\
 \to \left\{ \begin{array}{l}
\left( {2{m^2} - 1} \right)y = m + 3\\
x = 2my - 1
\end{array} \right.\\
 \to \left\{ \begin{array}{l}
y = \dfrac{{m + 3}}{{2{m^2} - 1}}\\
x = 2m.\dfrac{{m + 3}}{{2{m^2} - 1}} - 1
\end{array} \right.\\
 \to \left\{ \begin{array}{l}
y = \dfrac{{m + 3}}{{2{m^2} - 1}}\\
x = \dfrac{{2{m^2} + 6m - 2{m^2} + 1}}{{2{m^2} - 1}}
\end{array} \right.\\
 \to \left\{ \begin{array}{l}
y = \dfrac{{m + 3}}{{2{m^2} - 1}}\\
x = \dfrac{{6m + 1}}{{2{m^2} - 1}}
\end{array} \right.\\
a)Thay:m = 1\\
 \to \left\{ \begin{array}{l}
y = \dfrac{{1 + 3}}{{{{2.1}^2} - 1}}\\
x = \dfrac{{6.1 + 1}}{{{{2.1}^2} - 1}}
\end{array} \right.\\
 \to \left\{ \begin{array}{l}
y = 4\\
x = 7
\end{array} \right.\\
b)DK:2{m^2} - 1 \ne 0\\
 \to {m^2} \ne \dfrac{1}{2}\\
 \to m \ne  \pm \dfrac{1}{{\sqrt 2 }}\\
B6:\\
\left\{ \begin{array}{l}
2x + y = 5m - 1\\
x - 2y = 2
\end{array} \right.\\
 \to \left\{ \begin{array}{l}
x = 2y + 2\\
2\left( {2y + 2} \right) + y = 5m - 1
\end{array} \right.\\
 \to \left\{ \begin{array}{l}
x = 2y + 2\\
5y = 5m - 5
\end{array} \right.\\
 \to \left\{ \begin{array}{l}
y = m - 1\\
x = 2\left( {m - 1} \right) + 2
\end{array} \right.\\
 \to \left\{ \begin{array}{l}
y = m - 1\\
x = 2m
\end{array} \right.\\
a)Thay:m = 1\\
 \to \left\{ \begin{array}{l}
x = 2.1\\
y = 1 - 1
\end{array} \right.\\
 \to \left\{ \begin{array}{l}
x = 2\\
y = 0
\end{array} \right.\\
b)Do:{x^2} - 2{y^2} = 1\\
 \to 4{m^2} - 2\left( {{m^2} - 2m + 1} \right) = 1\\
 \to 2{m^2} + 4m - 3 = 0\\
 \to \left[ \begin{array}{l}
m = \dfrac{{ - 2 + \sqrt {10} }}{2}\\
m = \dfrac{{ - 2 - \sqrt {10} }}{2}
\end{array} \right.
\end{array}\)