Đáp án:
2.B
Do AM<AN<AC
Nên theo Pytago ta có:
$\begin{array}{l}
\sqrt {A{B^2} + A{M^2}} < \sqrt {A{B^2} + A{N^2}} < \sqrt {A{B^2} + A{C^2}} \\
\Rightarrow \sqrt {B{M^2}} < \sqrt {B{N^2}} < \sqrt {B{C^2}} \\
\Rightarrow BM < BN < BC
\end{array}$
3.A
a)DO AB> AC nên HB > HC
$\begin{array}{l}
\Rightarrow H{B^2} + D{H^2} > H{C^2} + D{H^2}\\
\Rightarrow D{B^2} > D{C^2}\\
\Rightarrow DB > DC
\end{array}$
b)
$\begin{array}{l}
MD < MA\\
\Rightarrow M{D^2} + M{B^2} < M{A^2} + M{B^2}\\
\Rightarrow D{B^2} < A{B^2}\\
\Rightarrow DB < AB
\end{array}$
3.B
$\begin{array}{l}
MP > MN\\
\Rightarrow KP > KN\\
\Rightarrow K{P^2} + Q{K^2} > K{N^2} + Q{K^2}\\
\Rightarrow Q{P^2} > Q{N^2}\\
\Rightarrow QP > QN
\end{array}$