Đáp án:
\(\begin{array}{l}
B23:\\
a)B5 = \dfrac{{2 - 5\sqrt x }}{{\sqrt x + 3}}\\
b)x = \dfrac{1}{{121}}\\
B24:\\
a)B6 = \dfrac{{\sqrt x - 2}}{{\sqrt x + 1}}\\
b)0 \le x < 4
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
B23:\\
a)DK:x \ge 0;x \ne 1\\
B5 = \dfrac{{15\sqrt x - 11 - \left( {3\sqrt x - 2} \right)\left( {\sqrt x + 3} \right) - \left( {2\sqrt x + 3} \right)\left( {\sqrt x - 1} \right)}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 3} \right)}}\\
= \dfrac{{15\sqrt x - 11 - 3x - 7\sqrt x + 6 - 2x - \sqrt x + 3}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 3} \right)}}\\
= \dfrac{{ - 5x + 7\sqrt x - 2}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 3} \right)}}\\
= \dfrac{{\left( {\sqrt x - 1} \right)\left( {2 - 5\sqrt x } \right)}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 3} \right)}}\\
= \dfrac{{2 - 5\sqrt x }}{{\sqrt x + 3}}\\
b)B5 = \dfrac{1}{2}\\
\to \dfrac{{2 - 5\sqrt x }}{{\sqrt x + 3}} = \dfrac{1}{2}\\
\to 4 - 10\sqrt x = \sqrt x + 3\\
\to 11\sqrt x = 1\\
\to \sqrt x = \dfrac{1}{{11}}\\
\to x = \dfrac{1}{{121}}\\
B24:\\
a)DK:x \ge 0;x \ne \left\{ {4;9} \right\}\\
B6 = \dfrac{{1 + \sqrt x - \sqrt x }}{{\sqrt x + 1}}:\dfrac{{\left( {\sqrt x + 3} \right)\left( {\sqrt x - 3} \right) - \left( {\sqrt x + 2} \right)\left( {\sqrt x - 2} \right) + \sqrt x + 2}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x - 3} \right)}}\\
= \dfrac{1}{{\sqrt x + 1}}.\dfrac{{\left( {\sqrt x - 2} \right)\left( {\sqrt x - 3} \right)}}{{x - 9 - x + 4 + \sqrt x + 2}}\\
= \dfrac{1}{{\sqrt x + 1}}.\dfrac{{\left( {\sqrt x - 2} \right)\left( {\sqrt x - 3} \right)}}{{\sqrt x - 3}}\\
= \dfrac{{\sqrt x - 2}}{{\sqrt x + 1}}\\
b)B6 < 0\\
\to \dfrac{{\sqrt x - 2}}{{\sqrt x + 1}} < 0\\
\to \sqrt x - 2 < 0\\
\to 0 \le x < 4
\end{array}\)
