a.\(\Delta ABC\) có : AD là p/g trong góc A \(\Rightarrow\dfrac{DB}{DC}=\dfrac{AB}{AC}=\dfrac{8}{6}=\dfrac{4}{3}\)

b.\(\Delta AHB\) và \(\Delta CHA\) có : góc H chung ; \(\widehat{HAB}=\widehat{C}\) ( cùng phụ \(\widehat{HAC}\) )

=> 2 \(\Delta\) đồng dạng \(\Rightarrow\dfrac{S_{AHB}}{S_{CHA}}=\left(\dfrac{AB}{AC}\right)^2=\left(\dfrac{8}{6}\right)^2=\dfrac{16}{9}\)

1.B 

2.B

3.D 

Whose blue bag is it ? 

\(M_N=11,25.2=22,5< 28=M_{C_2H_4}\) => Ankan X là : \(CH_4\)

Gọi CTPT của anken Y là : \(C_nH_{2n}\left(n\ge2\right)\)

\(m_N=0,2.22,5=4,5\left(g\right)\)

\(n_{CO_2}=0,3\left(mol\right)=n_C\) \(\Rightarrow\)  \(m_C=0,3.12=3,6\left(g\right)\)

\(m_H=4,5-3,6=0,9\left(g\right)\Rightarrow n_{H_2O}=0,45\left(mol\right)\)        

\(n_X=n_{CO_2}-n_{H_2O}=0,45-0,3=0,15\left(mol\right)\)

\(n_Y=0,2-0,15=0,05\left(mol\right)\)

BT C : \(0,3=0,15.1+0,05n\Rightarrow n=3\)

Anken(Y) là : \(C_3H_6\)