Ta có:
\(D=\dfrac{1.2+2.4+3.6+4.8+5.10}{3.4+6.8+9.12+12.16+15.20}\\ =\dfrac{1.2+1.2.4+1.2.3.3+1.2.2.8+1.2.5.5}{3.4+3.4.2.2+3.4.3.3+3.4.2.8+3.4.5.5}\\ =\dfrac{1.2\left(4+3^2+2.8+5^2\right)}{3.4\left(2^2+3^2+2.8+5^2\right)}\\ =\dfrac{1.2}{3.4}=\dfrac{1}{6}\)
Vậy \(D=\dfrac{1}{6}\)
so sánh
A = 1.2 + 2.4 + 3.6 + 4.8 + 5.10 / 3.4 + 6.8 + 9.12 + 12. 16 + 15. 20
B = 111111/ 666665
1,Tính
a, A = \(\dfrac{1}{1.2}\) + \(\dfrac{1}{2.3}\) + \(\dfrac{1}{3.4}\) + . . . + \(\dfrac{1}{2017.2018}\)
b,B = \(\dfrac{5}{2.4}\) + \(\dfrac{5}{4.6}\) + \(\dfrac{5}{6.8}\) + . . . + \(\dfrac{5}{2016.2018}\)
c. C = \(\dfrac{1}{18}\) + \(\dfrac{1}{54}\) + \(\dfrac{1}{108}\) + . . . + \(\dfrac{1}{990}\)
D= \(\dfrac{3}{2.4}+\dfrac{3}{4.6}+\dfrac{3}{6.8}+...+\dfrac{3}{98.100}\)
GIÚP MIK GẤP
Tính: \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{2017.2018}+\dfrac{1}{2018.2019}\)
Tính tổng
=\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}\)
Cho A=\(\dfrac{1}{1.2}+\dfrac{1}{3.4}+\dfrac{1}{5.6}+...+\dfrac{1}{59.60}\) Chứng tỏ A > \(\dfrac{7}{12}\)
Chứng minh rằng:
\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}+\dfrac{1}{8.9}+\dfrac{1}{9.10}\)
Tính tổng M=3/2.4+3/4.6+3/6.8+....+3/100.102
tính nhanh
A=\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}\)
B=\(\dfrac{1}{10}+\dfrac{1}{15}+\dfrac{1}{21}+...+\dfrac{1}{120}\)
các bn giúp mk nha