giúp em vs ạ em hứa vote ạ Question giúp em vs ạ em hứa vote ạ in progress 0 Môn Toán Ladonna 4 years 2020-10-29T08:15:34+00:00 2020-10-29T08:15:34+00:00 1 Answers 75 views 0
Answers ( )
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
a,\\
A = \left( {\dfrac{{\sqrt x – 1}}{{3\sqrt x – 1}} – \dfrac{1}{{3\sqrt x + 1}} + \dfrac{{8\sqrt x }}{{9x – 1}}} \right):\left( {1 – \dfrac{{3\sqrt x – 2}}{{3\sqrt x + 1}}} \right)\\
= \left( {\dfrac{{\sqrt x – 1}}{{3\sqrt x – 1}} – \dfrac{1}{{3\sqrt x + 1}} + \dfrac{{8\sqrt x }}{{\left( {3\sqrt x – 1} \right)\left( {3\sqrt x + 1} \right)}}} \right):\dfrac{{\left( {3\sqrt x + 1} \right) – \left( {3\sqrt x – 2} \right)}}{{3\sqrt x + 1}}\\
= \dfrac{{\left( {\sqrt x – 1} \right)\left( {3\sqrt x + 1} \right) – \left( {3\sqrt x – 1} \right) + 8\sqrt x }}{{\left( {3\sqrt x – 1} \right)\left( {3\sqrt x + 1} \right)}}:\dfrac{3}{{3\sqrt x + 1}}\\
= \dfrac{{\left( {3x – 2\sqrt x – 1} \right) – 3\sqrt x + 1 + 8\sqrt x }}{{\left( {3\sqrt x – 1} \right)\left( {3\sqrt x + 1} \right)}}.\dfrac{{3\sqrt x + 1}}{3}\\
= \dfrac{{3x + 3\sqrt x }}{{3\sqrt x – 1}}.\dfrac{1}{3}\\
= \dfrac{{x + \sqrt x }}{{3\sqrt x – 1}}\\
x = 6 + 2\sqrt 5 = 5 + 2.\sqrt 5 .1 + 1 = {\left( {\sqrt 5 + 1} \right)^2}\\
\Rightarrow \sqrt x = \sqrt 5 + 1\\
\Rightarrow A = \dfrac{{\left( {6 + 2\sqrt 5 } \right) + \left( {\sqrt 5 + 1} \right)}}{{3.\left( {\sqrt 5 + 1} \right) – 1}} = \dfrac{{7 + 3\sqrt 5 }}{{3\sqrt 5 + 2}}\\
b,\\
A = \dfrac{6}{5} \Leftrightarrow \dfrac{{x + \sqrt x }}{{3\sqrt x – 1}} = \dfrac{6}{5}\\
\Leftrightarrow 5.\left( {x + \sqrt x } \right) = 6.\left( {3\sqrt x – 1} \right)\\
\Leftrightarrow 5x + 5\sqrt x = 18\sqrt x – 6\\
\Leftrightarrow 5x – 13\sqrt x + 6 = 0\\
\Leftrightarrow \left( {\sqrt x – 2} \right)\left( {5\sqrt x – 3} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\sqrt x = 2\\
\sqrt x = \dfrac{3}{5}
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 4\\
x = \dfrac{9}{{25}}
\end{array} \right.
\end{array}\)