Đáp án:
\(\begin{array}{l}
B2:\\
a)\dfrac{{\sqrt x + 1}}{{\sqrt x - 3}}\\
b)x = 4\\
c)x < 9\\
d)\left[ \begin{array}{l}
x = 49\\
x = 25\\
x = 16\\
x = 1
\end{array} \right.
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
B2:\\
a)DK:x \ge 0;x \ne \left\{ {4;9} \right\}\\
A = \dfrac{{2\sqrt x - 9}}{{x - 5\sqrt x + 6}} - \dfrac{{\sqrt x + 3}}{{\sqrt x - 2}} - \dfrac{{2\sqrt x + 1}}{{3 - \sqrt x }}\\
= \dfrac{{2\sqrt x - 9 - \left( {\sqrt x + 3} \right)\left( {\sqrt x - 3} \right) + \left( {2\sqrt x + 1} \right)\left( {\sqrt x - 2} \right)}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x - 3} \right)}}\\
= \dfrac{{2\sqrt x - 9 - x + 9 + 2x - 3\sqrt x - 2}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x - 3} \right)}}\\
= \dfrac{{x - \sqrt x - 2}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x - 3} \right)}}\\
= \dfrac{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 1} \right)}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x - 3} \right)}}\\
= \dfrac{{\sqrt x + 1}}{{\sqrt x - 3}}\\
b)A = - 3\\
\to \dfrac{{\sqrt x + 1}}{{\sqrt x - 3}} = - 3\\
\to \sqrt x + 1 = - 3\sqrt x + 9\\
\to 4\sqrt x = 8\\
\to \sqrt x = 2\\
\to x = 4\\
c)A < 1\\
\to \dfrac{{\sqrt x + 1}}{{\sqrt x - 3}} < 1\\
\to \dfrac{{\sqrt x + 1 - \sqrt x + 3}}{{\sqrt x - 3}} < 0\\
\to \dfrac{4}{{\sqrt x - 3}} < 0\\
\to \sqrt x - 3 < 0\\
\to x < 9\\
d)A = \dfrac{{\sqrt x - 3 + 4}}{{\sqrt x - 3}}\\
= 1 + \dfrac{4}{{\sqrt x - 3}}\\
A \in Z \to \dfrac{4}{{\sqrt x - 3}} \in Z\\
\to \sqrt x - 3 \in U\left( 4 \right)\\
\to \left[ \begin{array}{l}
\sqrt x - 3 = 4\\
\sqrt x - 3 = 2\\
\sqrt x - 3 = 1\\
\sqrt x - 3 = - 1\\
\sqrt x - 3 = - 2\\
\sqrt x - 3 = - 4\left( l \right)
\end{array} \right. \to \left[ \begin{array}{l}
x = 49\\
x = 25\\
x = 16\\
x = 4\left( l \right)\\
x = 1
\end{array} \right.
\end{array}\)
