Đáp án:
b) \(x = \dfrac{1}{{36}}\)
Giải thích các bước giải:
\(\begin{array}{l}
A = 2.2\sqrt 2 - 5\sqrt 2 + \sqrt 2 + 3\\
= 3\\
B = \left[ {\dfrac{{x - 1}}{{\sqrt x \left( {\sqrt x - 1} \right)}}} \right].\dfrac{1}{{\sqrt x - 1}}\\
= \dfrac{{x - 1}}{{\sqrt x \left( {\sqrt x - 1} \right)}}.\dfrac{1}{{\sqrt x - 1}}\\
= \dfrac{{\sqrt x - 1}}{{\sqrt x \left( {\sqrt x - 1} \right)}}\\
b)A = 2B\\
\to \dfrac{{\sqrt x - 1}}{{\sqrt x \left( {\sqrt x - 1} \right)}} = 2.3\\
\to \sqrt x - 1 = 6x - 6\sqrt x \\
\to 6x - 7\sqrt x + 1 = 0\\
\to \left( {\sqrt x - 1} \right)\left( {6\sqrt x - 1} \right) = 0\\
\to \left[ \begin{array}{l}
x = 1\left( l \right)\\
x = \dfrac{1}{{36}}
\end{array} \right.
\end{array}\)
