Đáp án:
$\begin{array}{l}
x > 0;x \ne 1;x \ne 4\\
B = \left( {1 - \dfrac{{4\sqrt x }}{{x - 1}} + \dfrac{1}{{x - 1}}} \right):\dfrac{{x - 2\sqrt x }}{{x - 1}}\\
= \dfrac{{x - 1 - 4\sqrt x + 1}}{{x - 1}}.\dfrac{{x - 1}}{{x - 2\sqrt x }}\\
= \dfrac{{x - 4\sqrt x }}{{x - 2\sqrt x }}\\
= \dfrac{{\sqrt x - 4}}{{\sqrt x - 2}}\\
B = \dfrac{1}{2}\\
\Rightarrow \dfrac{{\sqrt x - 4}}{{\sqrt x - 2}} = \dfrac{1}{2}\\
\Rightarrow 2\left( {\sqrt x - 4} \right) = \sqrt x - 2\\
\Rightarrow 2\sqrt x - 8 = \sqrt x - 2\\
\Rightarrow \sqrt x = 6\\
= x = 36\left( {tmdk} \right)\\
Vậy\,x = 36\\
B2)\\
M\left( {{x_M};{y_M}} \right)\\
Do:{d_{M - {\rm{Ox}}}} = \left| {{y_M}} \right| = 6 \Rightarrow \left[ \begin{array}{l}
{y_M} = 6\\
{y_M} = - 6
\end{array} \right.\\
+ Khi:{y_M} = 6 \Rightarrow 6 = 2{x_M} - 1 \Rightarrow {x_M} = \dfrac{7}{2}\\
+ Khi:{y_M} = - 6 \Rightarrow - 6 = 2{x_M} - 1 \Rightarrow {x_M} = - \dfrac{5}{2}\\
\Rightarrow M\left( {\dfrac{7}{2};6} \right)/M\left( { - \dfrac{5}{2}; - 6} \right)
\end{array}$
