5/7 + 4/9 = 73/63

4/5 - 2/3 = 2/15

9/11 + 3/8 = 105/88

16/25 - 2/5 = 6/25.

\(\dfrac{4}{3}:\dfrac{2}{5}=\dfrac{4}{3}\cdot\dfrac{5}{2}=\dfrac{20}{6}=\dfrac{10}{3}\)

\(\dfrac{1}{8}:7=\dfrac{1}{8\cdot7}=\dfrac{1}{56}\)

\(3:\dfrac{4}{5}=3\cdot\dfrac{5}{4}=\dfrac{15}{4}\)

\(\dfrac{4}{5}:3=\dfrac{4}{5\cdot3}=\dfrac{4}{15}\)

a.\(y:\dfrac{4}{5}=\dfrac{11}{8}\)

\(y=\dfrac{11}{8}\times\dfrac{4}{5}\)

\(y=\dfrac{11}{10}\)

b.\(\dfrac{11}{3}-y=\dfrac{1}{9}\)

\(y=\dfrac{11}{3}-\dfrac{1}{9}\)

\(y=\dfrac{32}{9}\)

c.\(\dfrac{1}{7}\times x=\dfrac{8}{5}\)

\(x=\dfrac{8}{5}:\dfrac{1}{7}\)

\(x=\dfrac{56}{5}\)

\(y=\dfrac{11}{8}.\dfrac{4}{5}\)

\(y=\dfrac{44}{40}\)

\(x=\dfrac{11}{3}-\dfrac{1}{9}\)

\(x=\dfrac{99-3}{27}=\dfrac{96}{27}\)

\(x=\dfrac{8}{5}:\dfrac{1}{7}\)

\(x=\dfrac{56}{5}\)

a)\(y=\dfrac{4}{5}\times\dfrac{11}{8}=\dfrac{44}{40}=1\dfrac{4}{40}\)

b)\(y=\dfrac{11}{3}-\dfrac{1}{9}=\dfrac{33}{9}-\dfrac{1}{9}=\dfrac{32}{9}\)

c)\(x=\dfrac{8}{5}\times7=\dfrac{56}{5}\)

3- 32/35

= 3/1-32/35

= 105/35 - 32/35

= 73/35

73/35

3 - \(\dfrac{4}{7}\) x \(\dfrac{8}{5}\) 

= 3 - \(\dfrac{32}{35}\)

= \(\dfrac{105}{35}\) - \(\dfrac{32}{35}\) 

= \(\dfrac{73}{35}\) 

5/8 x 6 + 1/3 = 15/4 + 1/3 = 49/12

5/4- 1/3 = 15/12 - 4/12 = 11/12

a) = 15/4 + 1/3 = 49/12

b) = 5/4 - 1/3 = 11/12

1. 

= 10 000 - 29 x 11

= 10 000 - 319

= 9681

2.

= ( 3/7 + 4/7 ) + ( 4/9 + 5/9 )

= 1 + 1

= 2

\(A=\dfrac{4}{1\cdot5}+\dfrac{4}{5\cdot9}+...+\dfrac{4}{2001\cdot2005}\)

\(A=1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-...+\dfrac{1}{2001}-\dfrac{1}{2005}\)

\(A=1-\dfrac{1}{2005}=\dfrac{2004}{2005}\)

\(B=\dfrac{3}{10\cdot12}+\dfrac{3}{12\cdot14}+...+\dfrac{3}{998\cdot1000}\)

\(\dfrac{2}{3}B=\dfrac{2}{10\cdot12}+...+\dfrac{2}{998\cdot1000}\)

\(\dfrac{2}{3}B=\dfrac{1}{10}-\dfrac{1}{12}+\dfrac{1}{12}-...+\dfrac{1}{998}-\dfrac{1}{1000}\)

\(\dfrac{2}{3}B=\dfrac{1}{10}-\dfrac{1}{1000}=\dfrac{99}{1000}\)

\(B=\dfrac{99}{1000}:\dfrac{2}{3}=\dfrac{297}{2000}\)

\(A=\dfrac{4}{1.5}+\dfrac{4}{5.9}+...+\dfrac{4}{2001.2005}\)

\(\Rightarrow A=4\left(\dfrac{1}{1.5}+\dfrac{1}{5.9}+...+\dfrac{1}{2001.2005}\right)\)

\(\Rightarrow A=4.\dfrac{1}{4}\left(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+...+\dfrac{1}{2001}-\dfrac{1}{2005}\right)\)

\(\Rightarrow A=1-\dfrac{1}{2005}\)

\(\Rightarrow A=\dfrac{2004}{2005}\)

\(B=\dfrac{3}{10.12}+\dfrac{3}{12.14}+...+\dfrac{3}{998.1000}\)

\(\Rightarrow B=3\left(\dfrac{1}{10.12}+\dfrac{1}{12.14}+...+\dfrac{1}{998.1000}\right)\)

\(\Rightarrow B=3.\dfrac{1}{2}\left(\dfrac{1}{10}-\dfrac{1}{12}+\dfrac{1}{12}-\dfrac{1}{12}+...+\dfrac{1}{998}-\dfrac{1}{1000}\right)\)

\(\Rightarrow B=\dfrac{3}{2}\left(\dfrac{1}{10}-\dfrac{1}{1000}\right)\)

\(\Rightarrow B=\dfrac{3}{2}.\dfrac{99}{1000}\)

\(\Rightarrow B=\dfrac{297}{2000}\)

D=115/132/103/132=115/103

\(D=\frac{\frac{88}{132}-\frac{33}{132}+\frac{60}{132}}{\frac{55}{132}+\frac{132}{132}-\frac{84}{132}}\)

\(D=\frac{\frac{115}{132}}{\frac{103}{132}}\)

\(D=\frac{115}{103}\)

D = \(\frac{115}{103}\)