Đáp án:
$\begin{array}{l}
1)Dkxd:x > 0;x \ne 4\\
x = 25\\
\Rightarrow \sqrt x = 5\\
\Rightarrow A = \dfrac{{5 + 10}}{5} = 3\\
2)B = \dfrac{{\sqrt x + 1}}{{\sqrt x - 2}} - \dfrac{{\sqrt x + 4}}{{x - 2\sqrt x }} + \dfrac{1}{{\sqrt x }}\\
= \dfrac{{\sqrt x \left( {\sqrt x + 1} \right) - \sqrt x - 4 + \sqrt x - 2}}{{\sqrt x \left( {\sqrt x - 2} \right)}}\\
= \dfrac{{x + \sqrt x - \sqrt x - 4 + \sqrt x - 2}}{{\sqrt x \left( {\sqrt x - 2} \right)}}\\
= \dfrac{{x + \sqrt x - 6}}{{\sqrt x \left( {\sqrt x - 2} \right)}}\\
= \dfrac{{\left( {\sqrt x + 3} \right)\left( {\sqrt x - 2} \right)}}{{\sqrt x \left( {\sqrt x - 2} \right)}}\\
= \dfrac{{\sqrt x + 3}}{{\sqrt x }}\\
3)M = A:B\\
= \dfrac{{\sqrt x + 10}}{{\sqrt x }}.\dfrac{{\sqrt x }}{{\sqrt x + 3}}\\
= \dfrac{{\sqrt x + 10}}{{\sqrt x + 3}}\\
= \dfrac{{\sqrt x + 3 + 7}}{{\sqrt x + 3}}\\
= 1 + \dfrac{7}{{\sqrt x + 3}}\\
\Rightarrow \sqrt x + 3 = 7\left( {do:\sqrt x + 3 \ge 3} \right)\\
\Rightarrow \sqrt x = 4\\
\Rightarrow x = 16\left( {tmdk} \right)\\
B3:\\
1)\left\{ \begin{array}{l}
{x^2} + x + \dfrac{4}{{y + 5}} = 6\\
3\left( {{x^2} + x} \right) - \dfrac{2}{{y + 5}} = 4
\end{array} \right.\\
Dat:\left\{ \begin{array}{l}
{x^2} + x = a\\
\dfrac{1}{{y + 5}} = b\left( {dk:y \ne - 5} \right)
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
a + 4b = 6\\
3a - 2b = 4
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
a + 4b = 6\\
6a - 4b = 8
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
7a = 14\\
a + 4b = 6
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
a = 2\\
b = 1
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
{x^2} + x = 2\\
\dfrac{1}{{y + 5}} = 1
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
{x^2} + x - 2 = 0\\
y + 5 = 1
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
\left( {x - 2} \right)\left( {x + 1} \right) = 0\\
y = - 4
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x = 2;y = - 4\\
x = - 1;y = - 4
\end{array} \right.\\
2)a)\Delta ' = {\left( {m + 1} \right)^2} - 2m\\
= {m^2} + 1 > 0\\
\Rightarrow pt\,luon\,co\,2\,nghiem\,phan\,biet.\\
b)Theo\,Viet:\left\{ \begin{array}{l}
{x_1} + {x_2} = 2\left( {m + 1} \right)\\
{x_1}{x_2} = 2m
\end{array} \right.\\
{x_1} - {x_2} = - 2\\
\Rightarrow {\left( {{x_1} - {x_2}} \right)^2} = 4\\
\Rightarrow {\left( {{x_1} + {x_2}} \right)^2} - 4{x_1}{x_2} = 4\\
\Rightarrow 4{\left( {m + 1} \right)^2} - 4.2m = 4
\end{array}$
$\begin{array}{l}
\Rightarrow {m^2} + 2m + 1 - 2m = 1\\
\Rightarrow {m^2} = 0\\
\Rightarrow m = 0
\end{array}$
Vậy m=0
