50pt thanks moi nguoi

Đáp án:

$\begin{array}{l}
a)m =  - 1\\
 \Rightarrow \left\{ \begin{array}{l}
 - 2x + y = 2\\
8x - y = 1
\end{array} \right.\\
 \Rightarrow \left\{ \begin{array}{l}
 - 2x + y + 8x - y = 2 + 1\\
y = 8x - 1
\end{array} \right.\\
 \Rightarrow \left\{ \begin{array}{l}
6x = 3\\
y = 8x - 1
\end{array} \right.\\
 \Rightarrow \left\{ \begin{array}{l}
x = \dfrac{1}{2}\\
y = 3
\end{array} \right.\\
b)\left( {x;y} \right) = \left( {2; - 6} \right)\\
 \Rightarrow \left\{ \begin{array}{l}
2.m.2 - 6 = 2\\
8.2 - m.6 = m + 2
\end{array} \right.\\
 \Rightarrow \left\{ \begin{array}{l}
m = 2\\
16 - 6m = m + 2
\end{array} \right.\\
 \Rightarrow \left\{ \begin{array}{l}
m = 2\\
7m = 14
\end{array} \right.\\
 \Rightarrow m = 2\\
c)\left\{ \begin{array}{l}
2mx + y = 2\\
8x + my = m + 2
\end{array} \right.\\
 \Rightarrow \left\{ \begin{array}{l}
2{m^2}x + my = 2m\\
8x + my = m + 2
\end{array} \right.\\
 \Rightarrow 2{m^2}x - 8x = 2m - m - 2\\
 \Rightarrow 2.\left( {{m^2} - 4} \right).x = m - 2\\
 \Rightarrow 2\left( {m - 2} \right)\left( {m + 2} \right).x = m - 2\\
 + Khi:m = 2\\
 \Rightarrow 0.x = 0\left( {tm} \right)
\end{array}$

=> hpt có vô số nghiệm

$\begin{array}{l}
 + Khi:m =  - 2\\
 \Rightarrow 0.x =  - 4\left( {ktm} \right)
\end{array}$

=> hpt vô nghiệm

$\begin{array}{l}
 + Khi:m \ne 2;m \ne  - 2\\
 \Rightarrow \left\{ \begin{array}{l}
x = \dfrac{1}{{2\left( {m + 2} \right)}}\\
y = 2mx - 2
\end{array} \right.\\
 \Rightarrow \left\{ \begin{array}{l}
x = \dfrac{1}{{2m + 2}}\\
y = \dfrac{m}{{m + 1}} - 2 = \dfrac{{ - m - 2}}{{m + 1}}
\end{array} \right.\\
d)m \ne 2;m \ne  - 2\\
x - y > 0\\
 \Rightarrow \dfrac{1}{{2m + 2}} - \dfrac{{ - m - 2}}{{m + 1}} > 0\\
 \Rightarrow \dfrac{{1 + 2m + 4}}{{2m + 2}} > 0\\
 \Rightarrow \dfrac{{2m + 5}}{{m + 1}} > 0\\
 \Rightarrow \left[ \begin{array}{l}
m >  - 1\\
m <  - \dfrac{5}{2}
\end{array} \right.\\
Vay\,\left[ \begin{array}{l}
m >  - 1;m \ne 2\\
m <  - \dfrac{5}{2}
\end{array} \right.\\
e)m \ne 2;m \ne  - 2\\
P = {y^2} - 2x\\
 = {\left( {\dfrac{{ - m - 2}}{{m + 1}}} \right)^2} - 2.\dfrac{1}{{2m + 2}}\\
 = {\left( { - 1 - \dfrac{1}{{m + 1}}} \right)^2} - \dfrac{1}{{m + 1}}\\
 = {\left( {\dfrac{1}{{m + 1}}} \right)^2} + 2\left( {\dfrac{1}{{m + 1}}} \right) + 1 - \left( {\dfrac{1}{{m + 1}}} \right)\\
 = {\left( {\dfrac{1}{{m + 1}}} \right)^2} + \left( {\dfrac{1}{{m + 1}}} \right) + 1\\
 = {\left( {\dfrac{1}{{m + 1}} + \dfrac{1}{2}} \right)^2} + \dfrac{3}{4} \ge \dfrac{3}{4}\\
 \Rightarrow P \ge \dfrac{3}{4}\\
 \Rightarrow GTNN:P = \dfrac{3}{4}\\
Khi:m =  - 3
\end{array}$