Đáp án:
$\begin{array}{l}
B1)\\
a)A = \left( {\dfrac{{\sqrt {14} - \sqrt 7 }}{{\sqrt 2 - 1}} + \dfrac{{\sqrt {15} - \sqrt 5 }}{{\sqrt 3 - 1}}} \right)\left( {\sqrt 7 - \sqrt 5 } \right)\\
= \left( {\dfrac{{\sqrt 7 \left( {\sqrt 2 - 1} \right)}}{{\sqrt 2 - 1}} + \dfrac{{\sqrt 5 \left( {\sqrt 3 - 1} \right)}}{{\sqrt 3 - 1}}} \right).\left( {\sqrt 7 - \sqrt 5 } \right)\\
= \left( {\sqrt 7 + \sqrt 5 } \right)\left( {\sqrt 7 - \sqrt 5 } \right)\\
= 7 - 5 = 2\\
B = \left( {\dfrac{2}{{\sqrt x - 1}} - \dfrac{2}{{\sqrt x + 1}}} \right):\dfrac{1}{{\sqrt x - 1}}\\
= \dfrac{{2\left( {\sqrt x + 1} \right) - 2\left( {\sqrt x - 1} \right)}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}.\left( {\sqrt x - 1} \right)\\
= \dfrac{4}{{\sqrt x + 1}}\\
b)B > A\\
\Rightarrow \dfrac{4}{{\sqrt x + 1}} > 2\\
\Rightarrow \sqrt x + 1 < 2\\
\Rightarrow \sqrt x < 1\\
\Rightarrow x < 1\\
Vay\,0 \le x < 1\\
B2)\\
1)y = x - 3\\
Khi:x = 0 \Rightarrow y = - 3\\
\Rightarrow \left( {0; - 3} \right);\left( { - 2;0} \right) \in \left( d \right)\\
\Rightarrow \left\{ \begin{array}{l}
- 3 = b\\
0 = - 2a + b
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
b = - 3\\
a = \dfrac{{ - 3}}{2}
\end{array} \right.\\
\Rightarrow y = - \dfrac{3}{2}x - 3\\
2)Đặt:\dfrac{1}{{x + 1}} = a;\dfrac{1}{{y - 1}} = b\\
\Rightarrow \left\{ \begin{array}{l}
5a + b = 10\\
a - 3b = 18
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
15a + 3b = 30\\
a - 3b = 18
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
16a = 48\\
b = 10 - 5a
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
a = 3\\
b = - 5
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
\dfrac{1}{{x + 1}} = 3\\
\dfrac{1}{{y - 1}} = - 5
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
x + 1 = - \dfrac{1}{3}\\
y - 1 = - \dfrac{1}{5}
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
x = \dfrac{{ - 4}}{3}\\
y = \dfrac{4}{5}
\end{array} \right.\\
C3)\\
a)m = 1\\
\Rightarrow {x^2} - x - 2 = 0\\
\Rightarrow \left( {x - 2} \right)\left( {x + 1} \right) = 0\\
\Rightarrow \left[ \begin{array}{l}
x = 2\\
x = - 1
\end{array} \right.\\
b)\left\{ \begin{array}{l}
\Delta > 0\\
\dfrac{{ - b}}{a} > 0\\
\dfrac{c}{a} > 0
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
{m^2} - 4.\left( {2m - 4} \right) > 0\\
m > 0\\
2m - 4 > 0
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
{m^2} - 8m + 16 > 0\\
m > 0\\
m > 2
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
{\left( {m - 4} \right)^2} > 0\\
m > 2
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
m > 2\\
m \ne 4
\end{array} \right.
\end{array}$
Vậy $m > 2;m \ne 4$ thì pt có 2 nghiệm phân biệt đều dương.
