Đáp án:
$\begin{array}{l}
15)\\
{x^2} - 2mx - m + 2 = 0\\
\Leftrightarrow \Delta ' > 0\\
\Leftrightarrow {m^2} + m - 2 > 0\\
\Leftrightarrow \left( {m - 1} \right)\left( {m + 2} \right) > 0\\
\Leftrightarrow \left[ \begin{array}{l}
m > 1\\
m < - 2
\end{array} \right.\\
m \in \left( { - \infty ; - 2} \right) \cup \left( {1; + \infty } \right)\\
\Leftrightarrow B\\
16)\sin a = \dfrac{3}{5}\\
\Leftrightarrow \cos a = \dfrac{4}{5}\\
\left( {do:{{\sin }^2}a + {{\cos }^2}a = 1} \right)\\
\cos a = \cos \left( {2.\dfrac{a}{2}} \right) = 1 - 2{\sin ^2}\left( {\dfrac{a}{2}} \right)\\
\Leftrightarrow 2{\sin ^2}\dfrac{a}{2} = 1 - \dfrac{4}{5} = \dfrac{1}{5}\\
\Leftrightarrow {\sin ^2}\dfrac{a}{2} = \dfrac{1}{{10}}\\
\Leftrightarrow \sin \dfrac{a}{2} = \dfrac{{\sqrt {10} }}{{10}}\\
\Leftrightarrow A\\
17)A{M^2} = \dfrac{{A{B^2} + A{C^2}}}{2} - \dfrac{{B{C^2}}}{4}\\
= \dfrac{{{7^2} + {8^2}}}{2} - \dfrac{{{5^2}}}{4}\\
= \dfrac{{201}}{4}\\
\Leftrightarrow AM = \dfrac{{\sqrt {201} }}{2}\\
\Leftrightarrow C
\end{array}$
