Đáp án:
\(\begin{array}{l}
B1:\\
a)\dfrac{{2 - 5\sqrt x }}{{\sqrt x + 3}}\\
b)x = \dfrac{1}{{121}}\\
c)P \le \dfrac{2}{3}\\
B2:\\
\left[ \begin{array}{l}
A = \sqrt 2 \\
A = \sqrt {4x - 2}
\end{array} \right.
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
B1:\\
a)P = \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 + 3} \right)\left( {\sqrt x - 1} \right)}}\\
= \dfrac{{15\sqrt x - 11 - 3x - 7\sqrt x + 6 - 2x - \sqrt x + 3}}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x - 1} \right)}}\\
= \dfrac{{ - 5x + 7\sqrt x - 2}}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x - 1} \right)}}\\
= \dfrac{{\left( {2 - 5\sqrt x } \right)\left( {\sqrt x - 1} \right)}}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x - 1} \right)}}\\
= \dfrac{{2 - 5\sqrt x }}{{\sqrt x + 3}}\\
b)P = \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}}\\
c)Xét:P - \dfrac{2}{3} = \dfrac{{2 - 5\sqrt x }}{{\sqrt x + 3}} - \dfrac{2}{3}\\
= \dfrac{{6 - 15\sqrt x - 2\sqrt x - 6}}{{\sqrt x + 3}}\\
= \dfrac{{ - 17\sqrt x }}{{\sqrt x + 3}}\\
Do:x \ge 0 \to \left\{ \begin{array}{l}
17\sqrt x \ge 0\\
\sqrt x + 3 > 0
\end{array} \right.\\
\to \dfrac{{17\sqrt x }}{{\sqrt x + 3}} \ge 0\\
\to - \dfrac{{17\sqrt x }}{{\sqrt x + 3}} \le 0\\
\to P \le \dfrac{2}{3}\\
B2:\\
DK:x \ge \dfrac{1}{2}\\
A = \sqrt {x + \sqrt {2x - 1} } - \sqrt {x - \sqrt {2x - 1} } \\
= \dfrac{{\sqrt {2x + 2\sqrt {2x - 1} } - \sqrt {2x - 2\sqrt {2x - 1} } }}{{\sqrt 2 }}\\
= \dfrac{{\sqrt {2x - 1 + 2\sqrt {2x - 1} .1 + 1} - \sqrt {2x - 1 - 2\sqrt {2x - 1} .1 + 1} }}{{\sqrt 2 }}\\
= \dfrac{{\sqrt {{{\left( {\sqrt {2x - 1} + 1} \right)}^2}} - \sqrt {{{\left( {\sqrt {2x - 1} - 1} \right)}^2}} }}{{\sqrt 2 }}\\
= \dfrac{{\left| {\sqrt {2x - 1} + 1} \right| - \left| {\sqrt {2x - 1} - 1} \right|}}{{\sqrt 2 }}\\
\to \left[ \begin{array}{l}
A = \dfrac{{\sqrt {2x - 1} + 1 - \left( {\sqrt {2x - 1} - 1} \right)}}{{\sqrt 2 }}\left( {DK:\sqrt {2x - 1} \ge 1} \right)\\
A = \dfrac{{\sqrt {2x - 1} + 1 + \left( {\sqrt {2x - 1} - 1} \right)}}{{\sqrt 2 }}\left( {DK:\sqrt {2x - 1} < 1} \right)
\end{array} \right.\\
\to \left[ \begin{array}{l}
A = \dfrac{2}{{\sqrt 2 }}\\
A = \dfrac{{2\sqrt {2x - 1} }}{{\sqrt 2 }}
\end{array} \right.\\
\to \left[ \begin{array}{l}
A = \sqrt 2 \\
A = \sqrt {4x - 2}
\end{array} \right.
\end{array}\)
