Đáp án:
$\begin{array}{l}
a){\left( {1 + \sqrt 2 } \right)^2} - {\left( {1 - \sqrt 2 } \right)^2}\\
= 1 + 2\sqrt 2 + 2 - \left( {1 - 2\sqrt 2 + 2} \right)\\
= 4\sqrt 2 \\
b)dkxd:x \ge 0;x \ne 1\\
A = \dfrac{{1 - x\sqrt x }}{{1 + \sqrt x + x}} - \dfrac{{x + \sqrt x }}{{1 + \sqrt x }} - \dfrac{{x - {x^2}}}{{x - 1}}\\
= \dfrac{{\left( {1 - \sqrt x } \right)\left( {1 + \sqrt x + x} \right)}}{{1 + \sqrt x + x}} - \dfrac{{\sqrt x \left( {1 + \sqrt x } \right)}}{{1 + \sqrt x }} + \dfrac{{x\left( {x - 1} \right)}}{{x - 1}}\\
= 1 - \sqrt x - \sqrt x + x\\
= x - 2\sqrt x + 1\\
= {\left( {\sqrt x - 1} \right)^2} \ge 0\\
Khi:x \ne 1\\
\Rightarrow A > 0\\
c)y = a.x + b\\
A\left( {1;\dfrac{3}{4}} \right) \in d\\
\Rightarrow \dfrac{3}{4} = a + b\\
\Rightarrow b = \dfrac{3}{4} - a\\
\Rightarrow y = a.x + \dfrac{3}{4} - a\\
Xet:{x^2} = a.x + \dfrac{3}{4} - a\\
\Rightarrow {x^2} - a.x + a - \dfrac{3}{4} = 0\\
\Rightarrow \Delta = 0\\
\Rightarrow {a^2} - 4a + 3 = 0\\
\Rightarrow \left( {a - 1} \right)\left( {a - 3} \right) = 0\\
\Rightarrow \left[ \begin{array}{l}
a = 1\\
a = 3
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
d:y = x - \dfrac{1}{4}\\
d:y = 3x - \dfrac{9}{4}
\end{array} \right.\\
d)NP:y = a.x + b\\
\Rightarrow \left\{ \begin{array}{l}
- 4 = b\\
5 = 3a + b
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
b = - 4\\
a = 3
\end{array} \right.\\
\Rightarrow NP:y = 3x - 4\\
\Rightarrow M \in NP\\
\Rightarrow m + 1 = 3.2m - 4\\
\Rightarrow m = 1
\end{array}$
Vậy m=1
