Đáp án:

c. m>3

Giải thích các bước giải:

\(\begin{array}{l}
a.Thay:m = \sqrt 2 \\
Hpt \to \left\{ \begin{array}{l}
x\sqrt 2  + 4y = 10 - \sqrt 2 \\
x + y\sqrt 2  = 4
\end{array} \right.\\
 \to \left\{ \begin{array}{l}
x = 4 - y\sqrt 2 \\
\sqrt 2 \left( {4 - y\sqrt 2 } \right) + 4y = 10 - \sqrt 2 \left( * \right)
\end{array} \right.\\
\left( * \right) \to 4\sqrt 2  - 2y + 4y = 10 - \sqrt 2 \\
 \to 2y = 10 - 5\sqrt 2 \\
 \to y = 5 - \dfrac{5}{{\sqrt 2 }}\\
 \to x = 9 - 5\sqrt 2 \\
b.\left\{ \begin{array}{l}
x = 4 - my\\
4m - {m^2}y + 4y = 10 - m
\end{array} \right.\\
 \to \left\{ \begin{array}{l}
\left( {4 - {m^2}} \right)y = 10 - 5m\\
x = 4 - my
\end{array} \right.\\
 \to \left\{ \begin{array}{l}
y = \dfrac{{5\left( {2 - m} \right)}}{{\left( {2 - m} \right)\left( {2 + m} \right)}}\\
x = 4 - my
\end{array} \right.\\
 \to \left\{ \begin{array}{l}
y = \dfrac{5}{{m + 2}}\\
x = \dfrac{{4m + 8 - 5m}}{{m + 2}}
\end{array} \right.\\
 \to \left\{ \begin{array}{l}
x = \dfrac{{8 - m}}{{m + 2}}\\
y = \dfrac{5}{{m + 2}}
\end{array} \right.\\
Có:y - 5x = 1\\
 \to \dfrac{5}{{m + 2}} - 5.\dfrac{{8 - m}}{{m + 2}} = 1\left( {DK:m \ne  - 2} \right)\\
 \to 5 - 40 + 5m = m + 2\\
 \to 4m = 37\\
 \to m = \dfrac{{37}}{4}\left( {TM} \right)\\
c.\left\{ \begin{array}{l}
x < 1\\
y > 0
\end{array} \right.\\
 \to \left\{ \begin{array}{l}
\dfrac{{8 - m}}{{m + 2}} < 1\\
\dfrac{5}{{m + 2}} > 0
\end{array} \right.\\
 \to \left\{ \begin{array}{l}
m + 2 > 0\\
\dfrac{{8 - m - m - 2}}{{m + 2}} < 0
\end{array} \right.\\
 \to \left\{ \begin{array}{l}
m >  - 2\\
6 - 2m < 0
\end{array} \right.\\
 \to \left\{ \begin{array}{l}
m >  - 2\\
m > 3
\end{array} \right.\\
 \to m > 3
\end{array}\)