a=1/3+1/99=34/99

Ta có: \(A=\frac{1}{3}.\frac{1}{7}+\frac{1}{7}.\frac{1}{11}+\frac{1}{11}.\frac{1}{15}+...+\frac{1}{95}.\frac{1}{99}\)

              \(=\frac{1}{4}.\left(\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+.....+\frac{1}{95}-\frac{1}{99}\right)\)

               \(=\frac{1}{4}.\left(\frac{1}{3}-\frac{1}{99}\right)\)

                \(=\frac{1}{4}.\frac{32}{99}=\frac{8}{99}\)

A = 1/4 x (1/3-1/7+1/7-1/11+1/11-1/15+..........+1/95-1/99)

= 1/4 x(1/3-1/99)

= 1/4 x 32/99

= 8/99

\(4A=\frac{4}{3.7}+...+\frac{4}{95.99}\)

\(4A=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{95}-\frac{1}{99}\)

\(4A=\frac{1}{3}-\frac{1}{99}=\frac{32}{99}\)

\(\Rightarrow A=\frac{8}{99}\)

\(A=\frac{1}{3.7}+\frac{1}{7.11}+...+\frac{1}{95.99}\)

\(A=\frac{1}{4}.\left(\frac{4}{3.7}+\frac{4}{7.11}+...+\frac{4}{95.99}\right)\)

\(A=\frac{1}{4}.\left(\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{95}-\frac{1}{99}\right)\)

\(A=\frac{1}{4}.\left(\frac{1}{3}-\frac{1}{99}\right)\)

\(A=\frac{1}{4}.\frac{32}{99}\)

\(A=\frac{8}{99}\)

\(A=\frac{1}{3}.\frac{1}{7}+\frac{1}{7}.\frac{1}{11}+\frac{1}{11}.\frac{1}{15}+...+\frac{1}{95}.\frac{1}{99}.\)

    \(=\frac{1}{3.7}+\frac{1}{7.11}+\frac{1}{11.15}+...+\frac{1}{95.99}\)

\(4A=\frac{4}{3.7}+\frac{4}{7.11}+\frac{4}{11.15}+...+\frac{4}{95.99}\)

\(4A=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{95}-\frac{1}{99}\)

\(4A=\frac{1}{3}-\frac{1}{99}\)\(=\frac{33-1}{99}\)\(=\frac{32}{99}\)

\(\Rightarrow A=\frac{32}{99}:4=\frac{32}{99}.\frac{1}{4}=\frac{8}{99}\)

Phần 1)Đầu tiên bạn nhân B với 1 phần 4 rồi tính đến đoạn gần cuối sẽ ra 1/3 - 1/35 rồi quy đòng rồi tính sẽ ra kêt quả cuối là 32/105 nha

Mình lười lắm nên chỉ help 1 phần thui nha sr

giúp đi mà

\(A=\dfrac{1}{5\times7}+\dfrac{1}{7\times9}+\dfrac{1}{9\times11}+...+\dfrac{1}{87\times89}\)

\(A=\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{11}+...+\dfrac{1}{87}-\dfrac{1}{89}\)

\(A=\dfrac{1}{5}-\left(\dfrac{1}{7}-\dfrac{1}{7}\right)-\left(\dfrac{1}{9}-\dfrac{1}{9}\right)-...-\left(\dfrac{1}{87}-\dfrac{1}{87}\right)-\dfrac{1}{89}\)

\(A=\dfrac{1}{5}-\dfrac{1}{89}\)

\(A=\dfrac{84}{445}\)

Vậy, `A=84/445.`

A = \(\dfrac{1}{5\times7}\) + \(\dfrac{1}{7\times9}\)+\(\dfrac{1}{9\times11}\)+...+\(\dfrac{1}{87\times89}\)

A = \(\dfrac{1}{2}\) \(\times\)(  \(\dfrac{2}{5\times7}\)+\(\dfrac{2}{7\times9}\)+\(\dfrac{2}{9\times11}\)+...+\(\dfrac{2}{87\times89}\))

A = \(\dfrac{1}{2}\) \(\times\) ( \(\dfrac{1}{5}\) - \(\dfrac{1}{7}\) + \(\dfrac{1}{7}\) - \(\dfrac{1}{9}\) + \(\dfrac{1}{11}\) +...+ \(\dfrac{1}{87}\) - \(\dfrac{1}{89}\))

A = \(\dfrac{1}{2}\) \(\times\) (\(\dfrac{1}{5}\) - \(\dfrac{1}{89}\))

A = \(\dfrac{1}{2}\) \(\times\) \(\dfrac{84}{445}\) 

A = \(\dfrac{42}{445}\)

\(B=\frac{1}{3}-\frac{3}{4}+0,6+\frac{1}{64}-\frac{2}{9}-\frac{1}{36}+\frac{1}{15}\)

\(\Rightarrow B=\frac{3}{15}-\frac{48}{64}+\frac{9}{15}+\frac{1}{64}-\frac{8}{36}-\frac{1}{36}+\frac{1}{15}\)

\(\Rightarrow B=\frac{3}{15}+\frac{9}{15}+\frac{1}{15}+\left(-\frac{48}{64}+\frac{1}{64}\right)+\left(-\frac{8}{36}-\frac{1}{36}\right)\)

\(\Rightarrow B=\frac{13}{15}-\frac{47}{64}-\frac{1}{4}\)

\(\Rightarrow B=-\frac{113}{960}\)

\(C=0\)

\(D=\frac{1}{99}-\frac{1}{99.98}-...-\frac{1}{3.2}-\frac{1}{2.1}\)

\(\Rightarrow D=\frac{1}{99}-\frac{1}{99}+\frac{1}{98}-\frac{1}{98}+...-\frac{1}{3}+\frac{1}{2}-\frac{1}{2}+1\)

\(\Rightarrow D=1\)

D= \(\frac{1}{99}-\frac{1}{99.98}-\frac{1}{98.97}......-\frac{1}{3.2}-\frac{1}{2.1}\)

=\(\frac{1}{99}-\left(\frac{1}{1.2}+\frac{1}{2.3}+.......+\frac{1}{97.98}+\frac{1}{98.99}\right)\)

=\(\frac{1}{99}-\left(1-\frac{1}{2}+\frac{1}{2}-.....-\frac{1}{98}-\frac{1}{99}\right)\)

=\(\frac{1}{99}-\left[1-(\frac{1}{2}-\frac{1}{2}+......+\frac{1}{98}-\frac{1}{99})\right]\)

=\(\frac{1}{99}-\left(1-0-0-.....-0-\frac{1}{99}\right)\)

=\(\frac{1}{99}-1-\frac{1}{99}\)

=1

A = 4/3 x 7 + 4/7 x 11 + 4/11 x 15 + .... + 4/95 x 99

A = 4/3 - 4/7 + 4/7 - 4/11 + 4/11 - 4/15 + ..... + 4/95 - 4/99

A = 4/3 - 4/99

A = 128/99

\(A=4\left(\frac{4}{3.7}+\frac{4}{7.11}+.......+\frac{4}{95.99}\right)\)

\(=4\left(\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+.......+\frac{1}{95}-\frac{1}{99}\right)\)

\(=4.\left(\frac{1}{3}-\frac{1}{99}\right)\)

\(=4.\left(\frac{33}{99}-\frac{1}{99}\right)\)

\(=4.\frac{32}{99}=\frac{128}{99}\)

\(A=\frac{4}{3}.\frac{4}{7}+\frac{4}{7}.\frac{4}{11}+...+\frac{4}{95}.\frac{4}{99}\)

\(A=4\left(\frac{1}{3}.\frac{1}{7}+\frac{1}{7}.\frac{1}{11}+...+\frac{1}{95}.\frac{1}{99}\right)\)

\(A=4\left(\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{95}-\frac{1}{99}\right)\)

\(A=4\left(\frac{1}{3}-\frac{1}{99}\right)\)

\(A=4.\frac{32}{99}\)

\(A=\frac{128}{99}\)

a, \(\frac{5}{11}\times\frac{7}{25}+\frac{15}{11}\times\frac{1}{5}\)

\(=\frac{5}{11}\times\frac{7}{25}+\frac{5}{11}\times\frac{3}{5}\)

\(=\frac{5}{11}\times\left(\frac{7}{25}+\frac{3}{5}\right)\)

\(=\frac{5}{11}\times\frac{22}{25}\)

\(=\frac{2}{5}\)

b, \(\frac{3}{7}\times\frac{25}{19}-\frac{1}{7}\times\frac{18}{19}\)

\(=\frac{1}{7}\times\frac{75}{19}-\frac{1}{7}\times\frac{18}{19}\)

\(=\frac{1}{7}\times\left(\frac{75}{19}-\frac{18}{19}\right)\)

\(=\frac{1}{7}\times3\)

\(=\frac{3}{7}\)