Đáp án:
\(\begin{array}{l}
B8:\\
1)A = \dfrac{1}{{13}}\\
2)B = \dfrac{{\sqrt x + 6}}{{\sqrt x - 1}}\\
3)0 \le x < 1\\
B9)\\
1)A = 1\\
2)B = \dfrac{{3\sqrt x + 1}}{{\sqrt x + 3}}\\
3)x = 0\\
B10:\\
1)A = 0\\
2)B = \dfrac{{\sqrt x - 3}}{2}
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
B8:\\
DK:x \ge 0;x \ne 1\\
1)Thay:x = \dfrac{1}{4}\\
\to A = \dfrac{{\sqrt {\dfrac{1}{4}} }}{{\sqrt {\dfrac{1}{4}} + 6}} = \dfrac{1}{{13}}\\
2)B = \dfrac{{4 + \left( {\sqrt x + 3} \right)\left( {\sqrt x - 1} \right) + 5\left( {\sqrt x + 1} \right)}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}\\
= \dfrac{{4 + x + 2\sqrt x - 3 + 5\sqrt x + 5}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}\\
= \dfrac{{x + 7\sqrt x + 6}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}\\
= \dfrac{{\left( {\sqrt x + 6} \right)\left( {\sqrt x + 1} \right)}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}\\
= \dfrac{{\sqrt x + 6}}{{\sqrt x - 1}}\\
3)P = A.B = \dfrac{{\sqrt x }}{{\sqrt x + 6}}.\dfrac{{\sqrt x + 6}}{{\sqrt x - 1}}\\
= \dfrac{{\sqrt x }}{{\sqrt x - 1}}\\
P < 0\\
\to \dfrac{{\sqrt x }}{{\sqrt x - 1}} < 0\\
\to \sqrt x - 1 < 0\left( {do:\sqrt x \ge 0\forall x \ge 0} \right)\\
\to 0 \le x < 1\\
B9)\\
DK:x \ge 0;x \ne 1\\
1)Thay:x = 4\\
\to A = \dfrac{{2\sqrt 4 + 1}}{{\sqrt 4 + 3}} = \dfrac{{2.2 + 1}}{{2 + 3}} = 1\\
2)B = \dfrac{{2x + 6 - 2\left( {\sqrt x - 1} \right) + \left( {\sqrt x - 3} \right)\left( {\sqrt x + 3} \right)}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 3} \right)}}\\
= \dfrac{{2x + 6 - 2\sqrt x + 2 + x - 9}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 3} \right)}}\\
= \dfrac{{3x - 2\sqrt x - 1}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 3} \right)}}\\
= \dfrac{{\left( {\sqrt x - 1} \right)\left( {3\sqrt x + 1} \right)}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 3} \right)}}\\
= \dfrac{{3\sqrt x + 1}}{{\sqrt x + 3}}\\
3)A + B = \dfrac{{2\sqrt x + 1}}{{\sqrt x + 3}} + \dfrac{{3\sqrt x + 1}}{{\sqrt x + 3}}\\
= \dfrac{{5\sqrt x + 2}}{{\sqrt x + 3}}\\
A + B \le 2\\
\to \dfrac{{5\sqrt x + 2}}{{\sqrt x + 3}} \le 2\\
\to \dfrac{{5\sqrt x + 2 - 2\sqrt x - 6}}{{\sqrt x + 3}} \le 0\\
\to \dfrac{{3\sqrt x - 4}}{{\sqrt x + 3}} \le 0\\
\to 3\sqrt x - 4 \le 0\left( {do:\sqrt x + 3 > 0\forall x \ge 0} \right)\\
\to 0 \le x \le \dfrac{{16}}{9};x \ne 1\\
Do:x \in Z \to x = 0\\
B10:\\
DK:x \ge 0;x \ne \left\{ {1;4} \right\}\\
1)Thay:x = 9\\
\to A = \dfrac{{\sqrt 9 - 3}}{{\sqrt 9 - 2}} = 0\\
2)B = \dfrac{{\sqrt x \left( {\sqrt x + 1} \right) - 5\left( {\sqrt x - 1} \right) - 2}}{{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right)}}.\dfrac{{\sqrt x + 1}}{2}\\
= \dfrac{{x + \sqrt x - 5\sqrt x + 5 - 2}}{{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right)}}.\dfrac{{\sqrt x + 1}}{2}\\
= \dfrac{{x - 4\sqrt x + 3}}{{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right)}}.\dfrac{{\sqrt x + 1}}{2}\\
= \dfrac{{\left( {\sqrt x - 3} \right)\left( {\sqrt x - 1} \right)}}{{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right)}}.\dfrac{{\sqrt x + 1}}{2}\\
= \dfrac{{\sqrt x - 3}}{{\sqrt x + 1}}.\dfrac{{\sqrt x + 1}}{2}\\
= \dfrac{{\sqrt x - 3}}{2}
\end{array}\)
