Đáp án:
c) \(\left[ \begin{array}{l}
x = 1\\
x = 2
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
a)Thay:x = 3 + 2\sqrt 2 \\
= 2 + 2\sqrt 2 .1 + 1\\
= {\left( {\sqrt 2 + 1} \right)^2}\\
\to A = \dfrac{{3 + 2\sqrt 2 - 2}}{{2 + \sqrt {{{\left( {\sqrt 2 + 1} \right)}^2}} }}\\
= \dfrac{{1 + 2\sqrt 2 }}{{3 + \sqrt 2 }} = \dfrac{{5\sqrt 2 - 1}}{7}\\
b)B = \left[ {\dfrac{{\left( {2\sqrt x - 1} \right)\left( {4x + 2\sqrt x + 1} \right)}}{{\sqrt x \left( {2\sqrt x - 1} \right)}} - \dfrac{{\left( {2\sqrt x + 1} \right)\left( {4x - 2\sqrt x + 1} \right)}}{{\sqrt x \left( {2\sqrt x + 1} \right)}}} \right].\dfrac{{2x - 1}}{{2x + 1}}\\
= \dfrac{{4x + 2\sqrt x + 1 - 4x + 2\sqrt x - 1}}{{\sqrt x }}.\dfrac{{2x - 1}}{{2x + 1}}\\
= \dfrac{{4\sqrt x }}{{\sqrt x }}.\dfrac{{2x - 1}}{{2x + 1}} = \dfrac{{8x - 4}}{{2x + 1}}\\
c)\dfrac{A}{B} = \dfrac{{x - 2}}{{4\sqrt x }}\\
\to \dfrac{{x - 2}}{{\sqrt x + 2}}:\dfrac{{8x - 4}}{{2x + 1}} = \dfrac{{x - 2}}{{4\sqrt x }}\\
\to \dfrac{{\left( {x - 2} \right)\left( {2x + 1} \right)}}{{\left( {\sqrt x + 2} \right)\left( {8x - 4} \right)}} = \dfrac{{x - 2}}{{4\sqrt x }}\\
\to \dfrac{{\left( {x - 2} \right)\left( {2x + 1} \right)}}{{\left( {\sqrt x + 2} \right)\left( {8x - 4} \right)}} - \dfrac{{x - 2}}{{4\sqrt x }} = 0\\
\to \left( {x - 2} \right)\left( {\dfrac{{2x + 1}}{{\left( {\sqrt x + 2} \right)\left( {8x - 4} \right)}} - \dfrac{1}{{4\sqrt x }}} \right) = 0\\
\to \left[ \begin{array}{l}
x = 2\\
\dfrac{{8x\sqrt x + 4\sqrt x - 8x\sqrt x - 16x + 4\sqrt x + 8}}{{4\sqrt x \left( {\sqrt x + 2} \right)\left( {8x - 4} \right)}} = 0
\end{array} \right.\\
\to \left[ \begin{array}{l}
- 16x + 8\sqrt x + 8 = 0\\
x = 2
\end{array} \right.\\
\to \left[ \begin{array}{l}
\sqrt x = 1\\
\sqrt x = - \dfrac{1}{2}\left( l \right)\\
x = 2
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 1\\
x = 2
\end{array} \right.
\end{array}\)
