Đáp án:
c) \(\left[ \begin{array}{l}
x = \dfrac{{31 - \sqrt {61} }}{{18}}\\
x = \dfrac{{7 + \sqrt {13} }}{{18}}
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
a)DK:x \ge 0;x \ne 1\\
C = \dfrac{{2\sqrt x + 2 - 2\sqrt x + 2 - 2\sqrt x }}{{2\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right)}}\\
= \dfrac{{ - 2\sqrt x + 4}}{{2\left( {x - 1} \right)}}\\
= \dfrac{{ - \sqrt x + 2}}{{x - 1}}\\
b)Thay:x = \dfrac{4}{9}\\
\to \sqrt x = \dfrac{2}{3}\\
\to C = \dfrac{{ - \dfrac{2}{3} + 2}}{{\dfrac{4}{9} - 1}} = \dfrac{4}{3}:\left( { - \dfrac{5}{9}} \right) = - \dfrac{{12}}{5}\\
c)\left| C \right| = 3\\
\to \left[ \begin{array}{l}
C = 3\\
C = - 3
\end{array} \right.\\
\to \left[ \begin{array}{l}
\dfrac{{ - \sqrt x + 2}}{{x - 1}} = 3\\
\dfrac{{ - \sqrt x + 2}}{{x - 1}} = - 3
\end{array} \right.\\
\to \left[ \begin{array}{l}
- \sqrt x + 2 = 3x - 3\\
- \sqrt x + 2 = - 3x + 3
\end{array} \right.\\
\to \left[ \begin{array}{l}
3x + \sqrt x - 5 = 0\\
3x - \sqrt x - 1 = 0
\end{array} \right.\\
\to \left[ \begin{array}{l}
\sqrt x = \dfrac{{ - 1 + \sqrt {61} }}{6}\\
\sqrt x = \dfrac{{ - 1 - \sqrt {61} }}{6}\left( l \right)\\
\sqrt x = \dfrac{{1 + \sqrt {13} }}{6}\\
\sqrt x = \dfrac{{1 - \sqrt {13} }}{6}\left( l \right)
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = \dfrac{{31 - \sqrt {61} }}{{18}}\\
x = \dfrac{{7 + \sqrt {13} }}{{18}}
\end{array} \right.
\end{array}\)
