Cho \(A=\frac{1}{3}+\frac{1}{7}+\frac{1}{11}+\frac{1}{15}+\frac{1}{19}+...+\frac{1}{99}\)
Tính A.
K
Khách
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20 tháng 5 2016
Ta có: \(A=\frac{1}{3}.\frac{1}{7}+\frac{1}{7}.\frac{1}{11}+\frac{1}{11}.\frac{1}{15}+...+\frac{1}{95}.\frac{1}{99}\)
\(=\frac{1}{4}.\left(\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+.....+\frac{1}{95}-\frac{1}{99}\right)\)
\(=\frac{1}{4}.\left(\frac{1}{3}-\frac{1}{99}\right)\)
\(=\frac{1}{4}.\frac{32}{99}=\frac{8}{99}\)
TK
21 tháng 5 2019
\(4A=\frac{4}{3.7}+...+\frac{4}{95.99}\)
\(4A=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{95}-\frac{1}{99}\)
\(4A=\frac{1}{3}-\frac{1}{99}=\frac{32}{99}\)
\(\Rightarrow A=\frac{8}{99}\)
21 tháng 5 2019
\(A=\frac{1}{3.7}+\frac{1}{7.11}+...+\frac{1}{95.99}\)
\(A=\frac{1}{4}.\left(\frac{4}{3.7}+\frac{4}{7.11}+...+\frac{4}{95.99}\right)\)
\(A=\frac{1}{4}.\left(\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{95}-\frac{1}{99}\right)\)
\(A=\frac{1}{4}.\left(\frac{1}{3}-\frac{1}{99}\right)\)
\(A=\frac{1}{4}.\frac{32}{99}\)
\(A=\frac{8}{99}\)
25 tháng 7 2019
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
6 tháng 6 2023
\(A=\dfrac{1}{5\times7}+\dfrac{1}{7\times9}+\dfrac{1}{9\times11}+...+\dfrac{1}{87\times89}\)
\(A=\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{11}+...+\dfrac{1}{87}-\dfrac{1}{89}\)
\(A=\dfrac{1}{5}-\left(\dfrac{1}{7}-\dfrac{1}{7}\right)-\left(\dfrac{1}{9}-\dfrac{1}{9}\right)-...-\left(\dfrac{1}{87}-\dfrac{1}{87}\right)-\dfrac{1}{89}\)
\(A=\dfrac{1}{5}-\dfrac{1}{89}\)
\(A=\dfrac{84}{445}\)
Vậy, `A=84/445.`
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
6 tháng 6 2023
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 = \(\dfrac{42}{445}\)
31 tháng 7 2019
\(B=\frac{1}{3}-\frac{3}{4}+0,6+\frac{1}{64}-\frac{2}{9}-\frac{1}{36}+\frac{1}{15}\)
\(\Rightarrow B=\frac{3}{15}-\frac{48}{64}+\frac{9}{15}+\frac{1}{64}-\frac{8}{36}-\frac{1}{36}+\frac{1}{15}\)
\(\Rightarrow B=\frac{3}{15}+\frac{9}{15}+\frac{1}{15}+\left(-\frac{48}{64}+\frac{1}{64}\right)+\left(-\frac{8}{36}-\frac{1}{36}\right)\)
\(\Rightarrow B=\frac{13}{15}-\frac{47}{64}-\frac{1}{4}\)
\(\Rightarrow B=-\frac{113}{960}\)
\(C=0\)
\(D=\frac{1}{99}-\frac{1}{99.98}-...-\frac{1}{3.2}-\frac{1}{2.1}\)
\(\Rightarrow D=\frac{1}{99}-\frac{1}{99}+\frac{1}{98}-\frac{1}{98}+...-\frac{1}{3}+\frac{1}{2}-\frac{1}{2}+1\)
\(\Rightarrow D=1\)
LT
11 tháng 8 2019
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
29 tháng 6 2016
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
29 tháng 6 2016
\(A=4\left(\frac{4}{3.7}+\frac{4}{7.11}+.......+\frac{4}{95.99}\right)\)
\(=4\left(\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+.......+\frac{1}{95}-\frac{1}{99}\right)\)
\(=4.\left(\frac{1}{3}-\frac{1}{99}\right)\)
\(=4.\left(\frac{33}{99}-\frac{1}{99}\right)\)
\(=4.\frac{32}{99}=\frac{128}{99}\)
13 tháng 6 2020
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, \(\frac{3}{7}\times\frac{25}{19}-\frac{1}{7}\times\frac{18}{19}\)
\(=\frac{1}{7}\times\frac{75}{19}-\frac{1}{7}\times\frac{18}{19}\)
\(=\frac{1}{7}\times\left(\frac{75}{19}-\frac{18}{19}\right)\)
\(=\frac{1}{7}\times3\)
\(=\frac{3}{7}\)
A=1.078688093