Đáp án:
$\begin{array}{l}
1a)\\
Dkxd:x \ge 0;x \ne 4;x \ne 9\\
M = \dfrac{{2\sqrt x - 9}}{{x - 5\sqrt x + 6}} + \dfrac{{2\sqrt x + 1}}{{\sqrt x - 3}} + \dfrac{{\sqrt x + 3}}{{2 - \sqrt x }}\\
= \dfrac{{2\sqrt x - 9 + \left( {2\sqrt x + 1} \right)\left( {\sqrt x - 2} \right) - \left( {\sqrt x + 3} \right)\left( {\sqrt x - 3} \right)}}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x - 2} \right)}}\\
= \dfrac{{2\sqrt x - 9 + 2x - 3\sqrt x - 2 - x + 9}}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x - 2} \right)}}\\
= \dfrac{{x - \sqrt x - 2}}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x - 2} \right)}}\\
= \dfrac{{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 2} \right)}}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x - 2} \right)}}\\
= \dfrac{{\sqrt x + 1}}{{\sqrt x - 3}}\\
b)x = 16\left( {tmdk} \right)\\
\Rightarrow \sqrt x = 4\\
\Rightarrow M = \dfrac{{4 + 1}}{{4 - 3}} = 5\\
c)M = \dfrac{5}{3}\\
\Rightarrow \dfrac{{\sqrt x + 1}}{{\sqrt x - 3}} = \dfrac{5}{3}\\
\Rightarrow 5\sqrt x - 15 = 3\sqrt x + 3\\
\Rightarrow 2\sqrt x = 18\\
\Rightarrow \sqrt x = 9\\
\Rightarrow x = 81\left( {tmdk} \right)\\
2)\left( d \right)//\left( {d'} \right)\\
\Rightarrow \left( d \right):y = - 2x + b\left( {b \ne 5} \right)\\
Khi:x = 0;y = 2020\\
\Rightarrow b = 2020\left( {tmdk} \right)\\
\Rightarrow \left( d \right):y = - 2x + 2020\\
3)\left\{ \begin{array}{l}
2x - y = m - 1\\
3x + y = 4m + 1
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
5x = 5m\\
y = 2x - m + 1
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
x = m\\
y = m + 1
\end{array} \right.\\
x + y > 1\\
\Rightarrow m + m + 1 > 1\\
\Rightarrow m > 0
\end{array}$
