Câu 1:

a: Xét ΔADC có ME//DC

nên \(\dfrac{AM}{MD}=\dfrac{AE}{EC}\)

b: Xét ΔCAB có EF//AB

nên \(\dfrac{CE}{EA}=\dfrac{CF}{FB}\)

=>\(\dfrac{AE}{EC}=\dfrac{BF}{FC}\)

c: ta có: \(\dfrac{AM}{MD}=\dfrac{AE}{EC}\)

\(\dfrac{AE}{EC}=\dfrac{BF}{FC}\)

Do đó: \(\dfrac{AM}{MD}=\dfrac{BF}{FC}\)

d: Ta có: \(\dfrac{AM}{MD}=\dfrac{BF}{FC}\)

=>\(\dfrac{AM+MD}{MD}=\dfrac{BF+FC}{FC}\)

=>\(\dfrac{AD}{MD}=\dfrac{BC}{FC}\)

=>\(\dfrac{DM}{DA}=\dfrac{CF}{CB}\)

Bài 2:

Xét ΔADC có OM//DC

nên \(\dfrac{OM}{DC}=\dfrac{AM}{AD}\)(1)

Xét ΔBDC có ON//DC

nên \(\dfrac{ON}{DC}=\dfrac{BN}{BC}\left(2\right)\)

Xét hình thang ABCD có MN//AB//CD

nên \(\dfrac{AM}{MD}=\dfrac{BN}{NC}\)

=>\(\dfrac{MD}{AM}=\dfrac{CN}{BN}\)

=>\(\dfrac{MD+AM}{AM}=\dfrac{CN+BN}{BN}\)

=>\(\dfrac{AD}{AM}=\dfrac{BC}{BN}\)

=>\(\dfrac{AM}{AD}=\dfrac{BN}{BC}\left(3\right)\)

Từ (1),(2),(3) suy ra OM=ON

\(Q=m\cdot c\cdot\Delta t=m\cdot c\cdot\left(t_2-t_1\right)\)

\(\Rightarrow t_2-t_1=\dfrac{Q}{m\cdot c}=\dfrac{13.68\cdot10^3}{0.3\cdot380}=120\left(^0C\right)\)

\(\Rightarrow t_2=120+20=140^0C\)

83.13 : 17= 1079 : 17 =  63\(\dfrac{8}{17}\)

Bài 2:

a: \(=\dfrac{3}{5}\cdot\dfrac{6}{7}=\dfrac{18}{35}\)

b: =3/5+8/25

=15/25+8/25

=23/25

Bài 3:

a: =>x=2/3*2/5=4/25

b: =>x=4/9*2/3=8/27

\(=3\left(\dfrac{1}{2}+\dfrac{1}{5}\right)=3\cdot\dfrac{7}{10}=\dfrac{21}{10}\)

3/2+3/5

=15/10+6/10

=21/10

\(\dfrac{3}{2}+\dfrac{3}{5}=\dfrac{15}{10}+\dfrac{6}{10}=\dfrac{21}{10}\)

\(=\dfrac{1}{4}\cdot\dfrac{3}{4}+\dfrac{1}{4}\cdot\dfrac{1}{4}=\dfrac{1}{4}\)

\(=\dfrac{1}{4}.\dfrac{3}{4}+\dfrac{1}{4}.\dfrac{1}{4}\)

\(=\dfrac{3}{16}+\dfrac{1}{16}\)

\(=\dfrac{4}{16}\)

\(=\dfrac{1}{4}\)

đổi 25 phần trăm thành 0,25 rồi tính như bình thường