GIẢI HẾT GIÚP MÌNH với đề trong hình đó

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