Đáp án:
C2:
b. \(\dfrac{2}{3}{\tan ^6}a\)
Giải thích các bước giải:
\(\begin{array}{l}
C1:\\
a.\left( {x - 5} \right)\left( {x - 2} \right) < 0\\
\to 2 < x < 5\\
b.DK:x \ne \left\{ {1;2} \right\}\\
\dfrac{{5x - 11 - 2{x^2} + 6x - 4}}{{\left( {x - 1} \right)\left( {x - 2} \right)}} \ge 0\\
\to \dfrac{{ - 2{x^2} + 11x - 15}}{{\left( {x - 1} \right)\left( {x - 2} \right)}} \ge 0\\
\to \dfrac{{\left( {3 - x} \right)\left( {2x - 5} \right)}}{{\left( {x - 1} \right)\left( {x - 2} \right)}} \ge 0
\end{array}\)
BXD:
x -∞ 1 2 5/2 3 +∞
y - // + // - 0 + 0 -
\(KL:x \in \left( {1;2} \right) \cup \left[ {\dfrac{5}{2};3} \right]\)
C2:
\(\begin{array}{l}
a.Do:a \in \left( {\dfrac{{3\pi }}{2};2\pi } \right)\\
\to \cos a > 0\\
Do:\sin a = - \dfrac{4}{{11}}\\
{\sin ^2}a + {\cos ^2}a = 1\\
\to \dfrac{{16}}{{121}} + {\cos ^2}a = 1\\
\to {\cos ^2}a = \dfrac{{105}}{{121}}\\
\to \cos a = \dfrac{{\sqrt {105} }}{{11}}\\
b.A = \dfrac{{2.\left( {\dfrac{{{{\sin }^2}a}}{{{{\cos }^2}a}} - {{\sin }^2}a} \right)}}{{3\left( {\dfrac{{{{\cos }^2}a}}{{{{\sin }^2}a}} - {{\cos }^2}a} \right)}}\\
= 2.\left( {\dfrac{{{{\sin }^2}a - {{\sin }^2}a.{{\cos }^2}a}}{{{{\cos }^2}a}}} \right):3\left( {\dfrac{{{{\cos }^2}a - {{\sin }^2}a.{{\cos }^2}a}}{{{{\sin }^2}a}}} \right)\\
= 2.\left( {\dfrac{{{{\sin }^2}a - {{\sin }^2}a.{{\cos }^2}a}}{{{{\cos }^2}a}}} \right).\dfrac{{{{\sin }^2}a}}{{3\left( {{{\cos }^2}a - {{\sin }^2}a.{{\cos }^2}a} \right)}}\\
= \dfrac{{2\left( {{{\sin }^4}a - {{\sin }^4}a.{{\cos }^2}a} \right)}}{{3\left( {{{\cos }^4}a - {{\sin }^2}a.{{\cos }^4}a} \right)}}\\
= \dfrac{{2{{\sin }^4}a\left( {1 - {{\cos }^2}a} \right)}}{{3{{\cos }^4}a\left( {1 - {{\sin }^2}a} \right)}}\\
= \dfrac{{2{{\sin }^4}a.{{\sin }^2}a}}{{3{{\cos }^4}a{{\cos }^2}a}} = \dfrac{{2{{\sin }^6}a}}{{3{{\cos }^6}a}}\\
= \dfrac{2}{3}{\tan ^6}a
\end{array}\)
