Các câu hỏi tương tự
\(\left(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+.....+\frac{1}{2017.2018}\right)-\left(\frac{1}{1010}+\frac{1}{1011}+\frac{1}{1012}+.....+\frac{1}{2017}\right)\)
cho \(a=\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+....+\frac{1}{2017.2018}\) ; \(b=\frac{1}{1010}+\frac{1}{1011}+\frac{1}{1012}+...+\frac{1}{2018}\) . Tính (a-b)^2019
â , tính M = \(\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right).......\left(1+\frac{1}{2017}\right)\left(1+\frac{1}{2018}\right)\)
b , Cho A = \(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+.....+\frac{1}{2017}-\frac{1}{2018}\)
c , B = \(\frac{1}{1010}+\frac{1}{1011}+.....+\frac{1}{2017}+\frac{1}{2018}.tinh\left(\frac{A}{B}\right)^{2018}\)
a)Afrac{1}{1.2}+frac{1}{2.3}+frac{1}{3.4}+...+frac{1}{99.100} 1b)Bfrac{1}{3}+left(frac{1}{3}right)^2+left(frac{1}{3}right)^3+...+left(frac{1}{3}right)^{100} frac{1}{2}c)frac{1}{1.2}+frac{1}{3.4}+frac{1}{5.6}+...+frac{1}{49.50}frac{1}{26}+frac{1}{27}+...+frac{1}{50}d)Afrac{1}{1.2}+frac{1}{3.4}+...+frac{1}{99.100}.CMRfrac{7}{12} A frac{5}{6}AI ĐÚNG MINK left(TICKright)CHO (làm đc trên 2 câu)
Đọc tiếp
a)A=\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}< 1\)
b)B=\(\frac{1}{3}+\left(\frac{1}{3}\right)^2+\left(\frac{1}{3}\right)^3+...+\left(\frac{1}{3}\right)^{100}< \frac{1}{2}\)
c)\(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{49.50}=\frac{1}{26}+\frac{1}{27}+...+\frac{1}{50}\)
d)A=\(\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{99.100}.CMR\frac{7}{12}< A< \frac{5}{6}\)
AI ĐÚNG MINK \(\left(TICK\right)\)CHO (làm đc trên 2 câu)
Tính \(\left(M-N\right)^2\)với \(M=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{2017}-\frac{1}{2018}+\frac{1}{2019}\)
và \(N=\frac{1}{1010}+\frac{1}{1011}+....+\frac{1}{2019}\)
Tập hợp các giá trị nguyên dương của x thỏa mãn:\(\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}\right)x<\frac{13}{7}\)có số phần tử là...........
tính B=\(\left(1-\frac{2}{2.3}\right)\)x\(\left(1-\frac{2}{3.4}\right)\)x.......x\(\left(1-\frac{2}{2017.2018}\right)\)
Cho A=\(\frac{1}{1011}\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{2019}\right)\)và B=\(\frac{1}{1010}\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{2020}\right)\)So sánh A và B
\(\left|x+\frac{1}{1.2}\right|+\left|x+\frac{1}{2.3}\right|+\left|x+\frac{1}{3.4}\right|+...+\left|x+\frac{1}{99.100}\right|=100x\)