Đáp án:
$\begin{array}{l}
2c)\\
{d_1} \bot d\\
\Rightarrow a.\left( { - 1} \right) = - 1\\
\Rightarrow a = 1\\
\Rightarrow {d_1}:y = x - m + 1\\
Xet:{x^2} = x - m + 1\\
\Rightarrow {x^2} - x + m - 1 = 0\\
\Rightarrow \Delta > 0\\
\Rightarrow 1 - 4m + 4 > 0\\
\Rightarrow m < \dfrac{5}{4}\\
6)AB = 6cm;\widehat C = {60^0}\\
\Rightarrow \tan \widehat C = \dfrac{{AB}}{{AC}}\\
\Rightarrow \tan {60^0} = \dfrac{6}{{AC}}\\
\Rightarrow AC = \dfrac{6}{{\sqrt 3 }} = 2\sqrt 3 \\
\Rightarrow BC = \sqrt {A{B^2} + A{C^2}} = 12\left( {cm} \right)\\
\Rightarrow TT:AM = \dfrac{{BC}}{2} = 6\left( {cm} \right)
\end{array}$
