\(A=\frac{4}{3.7}+\frac{4}{7.11}+....+\frac{4}{95.99}\)
\(=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{95}-\frac{1}{99}\)
\(=\frac{1}{3}-\frac{1}{99}\)
\(=\frac{32}{99}\)
TISNG BIỂU THỨC
A= 4/3.7+4/7.11+4/11.15+...................................................+4/95.99
tính tổng sau : \(A=\frac{4}{3.7}+\frac{4}{7.11}+\frac{4}{11.15}+...+\frac{4}{107.111}\)
Hãy tính 1 cách hợp lí:
\(\frac{1}{3.7}+\)\(\frac{1}{7.11}+\frac{1}{11.15}+......+\frac{1}{95.99}\)
Tính nhanh
A=\(\frac{4}{3.7}+\frac{4}{7.11}+\frac{4}{11.15}+...+\frac{1}{107.111}\)
tính tổng \(\frac{4}{3.7}+\frac{4}{7.11}+\frac{4}{11.15}+...+\frac{4}{1023.1027}\)
\(\frac{4}{3.7}+\frac{4}{7.11}+\frac{4}{11.15}+\frac{4}{15.19}+\frac{4}{19.23}+\frac{4}{23.27}\)
Cho A = \(\frac{4}{3.7}+\frac{4}{7.11}+\frac{4}{11.15}+...+\frac{4}{103.107}\). So sánh A với 1
Tìm x, biết:
\(\frac{1}{3.7}+\frac{1}{7.11}+\frac{1}{11.15}+...+\frac{1}{x\left(x+4\right)}=\frac{43}{552}\)
Chứng minh rằng:
a,\(\frac{5}{3.7}+\frac{5}{7.11}+\frac{5}{11.15}+...+\frac{5}{\left(4n-1\right).\left(4n+3\right)}=\frac{5n}{3.\left(4n+3\right)}\)
b,\(\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+...+\frac{1}{1+2+3+...+100}< \frac{1}{4}\)
