b: Xét tứ giác BFEC có 

\(\widehat{BFC}=\widehat{BEC}=90^0\)

Do đó: BFEC là tứ giác nội tiếp

Xét tứ giác AEIF có 

\(\widehat{AEI}+\widehat{AFI}=180^0\)

Do đó: AEIF là tứ giác nội tiếp

1=

3.1:

Ô tô đi hết 1h50'+40'=2h30'

3.2:

Ô tô đi hết:

17h50'-15h15'

=2h35'

`#040911`

`b)`

\(A=\dfrac{1}{1\times3}+\dfrac{1}{3\times5}+\dfrac{1}{5\times7}+...+\dfrac{1}{19\times21}\)

`=`\(\dfrac{1}{2}\times\left(\dfrac{2}{1\times3}+\dfrac{2}{3\times5}+\dfrac{2}{5\times7}+...+\dfrac{2}{19\times21}\right)\)

`=`\(\dfrac{1}{2}\times\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{19}-\dfrac{1}{21}\right)\)

`=`\(\dfrac{1}{2}\times\left(1-\dfrac{1}{21}\right)\)

\(=\dfrac{1}{2}\times\dfrac{20}{21}\\ =\dfrac{10}{21}\\ \text{ Vậy, A = }\dfrac{10}{21}\)

Câu 1.

Gọi \(\left\{{}\begin{matrix}n_{CH_4}=a\left(mol\right)\\n_{C_2H_6}=b\left(mol\right)\end{matrix}\right.\)

\(n_{CO_2}=\dfrac{6,72}{22,4}=0,3mol\)

\(CH_4+2O_2\underrightarrow{t^o}CO_2+2H_2O\)

\(C_2H_6+\dfrac{7}{2}O_2\underrightarrow{t^o}2CO_2+3H_2O\)

\(\Rightarrow\left\{{}\begin{matrix}16a+30b=4,6\\a+2b=0,3\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a=0,1\\b=0,1\end{matrix}\right.\)

\(\%m_{CH_4}=\dfrac{0,1\cdot16}{4,6}\cdot100\%=34,78\%\)

\(\%m_{C_2H_6}=100\%-34,785=65,22\%\)