Đáp án:
c) x=4
Giải thích các bước giải:
\(\begin{array}{l}
B8:\\
a)DK:x \ge 0;x \ne 1\\
C = \dfrac{{\sqrt x + 1 - \sqrt x + 1 - 2\sqrt x }}{{2\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right)}}\\
= \dfrac{{2 - 2\sqrt x }}{{2\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right)}}\\
= \dfrac{{1 - \sqrt x }}{{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right)}} = - \dfrac{1}{{\sqrt x + 1}}\\
b)Thay:x = \dfrac{4}{9}\\
\to C = - \dfrac{1}{{\sqrt {\dfrac{4}{9}} + 1}} = - \dfrac{3}{5}\\
c)\left| C \right| = \dfrac{1}{3}\\
\to \left[ \begin{array}{l}
C = \dfrac{1}{3}\\
C = - \dfrac{1}{3}
\end{array} \right. \to \left[ \begin{array}{l}
- \dfrac{1}{{\sqrt x + 1}} = \dfrac{1}{3}\\
- \dfrac{1}{{\sqrt x + 1}} = - \dfrac{1}{3}
\end{array} \right.\\
\to \left[ \begin{array}{l}
- 3 = \sqrt x + 1\\
\sqrt x + 1 = 3
\end{array} \right.\\
\to \left[ \begin{array}{l}
\sqrt x = - 4\left( l \right)\\
\sqrt x = 2
\end{array} \right.\\
\to x = 4
\end{array}\)
