Tính A=4/3×4/7+4/7×4/11+4/11×4/15+...+4/95×4/99
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
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
\(A=4\left(\frac{4}{3.7}+\frac{4}{7.11}+.......+\frac{4}{95.99}\right)\)
\(=4\left(\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+.......+\frac{1}{95}-\frac{1}{99}\right)\)
\(=4.\left(\frac{1}{3}-\frac{1}{99}\right)\)
\(=4.\left(\frac{33}{99}-\frac{1}{99}\right)\)
\(=4.\frac{32}{99}=\frac{128}{99}\)
Q
17 tháng 4 2017
\(\Rightarrow A=\frac{4}{3.7}+\frac{4}{7.11}+\frac{4}{11.15}+...+\frac{4}{95.99}\)
\(A=4\left(\frac{1}{3.7}+\frac{1}{7.11}+\frac{1}{11.15}+...+\frac{1}{95.99}\right)\)
\(A=4.\frac{1}{4}\left(\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{15}+...+\frac{1}{95}-\frac{1}{99}\right)\)
\(A=\frac{4}{4}\left(\frac{1}{3}-\frac{1}{99}\right)\)
\(A=\frac{32}{99}\)
17 tháng 4 2017
\(\frac{4}{3}.\frac{4}{7}+\frac{4}{7}.\frac{4}{11}+\frac{4}{11}.\frac{4}{15}+...+\frac{4}{95}.\frac{4}{99}\)
\(\Leftrightarrow\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{15}+...+\frac{1}{95}-\frac{1}{99}\)
\(=\frac{1}{3}-\frac{1}{99}=\frac{32}{99}\)
\(\Leftrightarrow A=\frac{32}{99}\)
19 tháng 3 2017
A = \(\dfrac{4}{3}\) . \(\dfrac{4}{7}\) + \(\dfrac{4}{7}\) . \(\dfrac{4}{11}\) + \(\dfrac{4}{11}\) . \(\dfrac{4}{15}\) + ... + \(\dfrac{4}{95}\) . \(\dfrac{4}{99}\)
A = \(\dfrac{4.4}{3.7}\) + \(\dfrac{4.4}{7.11}\) + \(\dfrac{4.4}{11.15}\) + ... + \(\dfrac{4.4}{95.99}\)
A = \(\dfrac{16}{3.7}\) + \(\dfrac{16}{7.11}\) + \(\dfrac{16}{11.15}\) + ... + \(\dfrac{16}{95.99}\)
A = 4.( \(\dfrac{4}{3.7}\) + \(\dfrac{4}{7.11}\) + \(\dfrac{4}{11.15}\) + ... + \(\dfrac{4}{95.99}\))
A = 4.( \(\dfrac{1}{3}\) - \(\dfrac{1}{7}\) + \(\dfrac{1}{7}\) - \(\dfrac{1}{11}\) + \(\dfrac{1}{11}\) - \(\dfrac{1}{15}\) + ... + \(\dfrac{1}{95}\) - \(\dfrac{1}{99}\))
A = 4.(\(\dfrac{1}{3}\) - \(\dfrac{1}{99}\))
A = 4.(\(\dfrac{33}{99}\) + \(\dfrac{-1}{99}\))
A = 4. \(\dfrac{32}{99}\)
A = \(\dfrac{4.32}{99}\)
A = \(\dfrac{128}{99}\)
Vậy A = \(\dfrac{128}{99}\)
TP
tính giá trị biểu thức:A=4 phần 3 x 4 phần 7+4 phần 7 x 4 phần 11 +...........+4 phần 95 x 4 phần 99
1
19 tháng 4 2017
4/3.4/7+4/7.4/11+...+4/95.4/99
=4/3.7+4/7.11+...+4/95.99
=1/3-1/7+1/7-1/11+...+1/95-1/99
=1/3-1/99
=32/99
NN
Nguyễn Ngọc Anh Minh
CTVHS
VIP
25 tháng 4 2017
\(\frac{A}{4}=\frac{4}{3.7}+\frac{4}{7.11}+...+\frac{4}{95.99}\)
\(\frac{A}{4}=\frac{7-3}{3.7}+\frac{11-7}{7.11}+...+\frac{99-95}{95.99}\)
\(\frac{A}{4}=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{95}-\frac{1}{99}=\frac{1}{3}-\frac{1}{99}=\frac{32}{99}\)
\(A=\frac{4.32}{99}\)
24 tháng 7 2018
\(4.A=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+....+\frac{1}{95}-\frac{1}{99}\\ 4.A=\frac{1}{3}-\frac{1}{99}\\ 4.A=\frac{32}{99}\\ A=\frac{32}{99}:4\\ A=\frac{8}{99}\)
\(A=\frac{4}{3}\cdot\frac{4}{7}+\frac{4}{7}\cdot\frac{4}{11}+\frac{4}{11}\cdot\frac{4}{15}+...+\frac{4}{95}\cdot\frac{4}{99}\)
\(A=\frac{16}{3\cdot7}+\frac{16}{7\cdot11}+\frac{16}{11\cdot15}+...+\frac{16}{95\cdot99}\)
\(A=4\cdot\left(\frac{4}{3\cdot7}+\frac{4}{7\cdot11}+\frac{4}{11\cdot15}+...+\frac{4}{95\cdot99}\right)\)
\(A=4\cdot\left(\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{15}+...+\frac{1}{95}-\frac{1}{99}\right)\)
\(A=4\cdot\left(\frac{1}{3}-\frac{1}{99}\right)\)
\(A=4\cdot\frac{32}{99}\)
\(A=\frac{128}{99}\)
\(A=\frac{4}{3}\times\frac{4}{7}+\frac{4}{7}\times\frac{4}{11}+...+\frac{4}{95}\times\frac{4}{99}\)
\(=4\times\frac{4}{3.7}+4\times\frac{4}{7.11}+...+4\times\frac{4}{95.99}\)
\(=4\times\left(\frac{4}{3.7}+\frac{4}{7.11}+\frac{4}{11.15}+...+\frac{4}{95.99}\right)\)
\(=4\times\left(\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{91}-\frac{1}{95}+\frac{1}{95}-\frac{1}{99}\right)\)
\(=4\times\left(\frac{1}{3}-\frac{1}{99}\right)\)
\(=4\times\frac{32}{99}\)
\(=\frac{128}{99}\)