Rút gọn:
a/ \(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2009.2000}\)
b/ \(B=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{1998.1999.2000}\)
c/ \(C=\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+...+\frac{1}{2006.2008}\)
CMR : a) 1/2! + 2/3! + 3/4! +...+ 99/100! < 1
b) 1.2-1/2! + 2.3-1/3! + 3.4-1/4! +...+ 99.100-1/100! < 2
Tính A = [1 + (1 + 2) + (1 + 2 + 3) + ..... + (1 + 2 + 3 + ..... + 98)]/(1.2 + 2.3 + 3.4 + ..... + 98.99)
Thuc hien phep tinh:
a/ \(\dfrac{5}{15}\)+ \(\dfrac{14}{15}\) -\(\dfrac{12}{9}\)+ \(\dfrac{2}{7}\)+\(\dfrac{11}{25}\) d/ (\(9\dfrac{3}{4}\):5.2+3.4.\(2\dfrac{7}{34}\)) :(\(-1\dfrac{9}{16}\))
e/ \(\dfrac{2^8.9^2}{6^4.8^2}\)
b/ 4.(\(-\dfrac{1}{2}\))^3+\(\dfrac{1}{2}\)
c/ \(\dfrac{10^3+5.10^2+5^3}{6^3+3.6^2+3^3}\)
bài 1 :
A = 1/ 1.2 + 1/3.4 + 1/5.6 + .........+ 1/ 2005 . 2006
B = 1/ 1004.2006 + 1/ 1005.2006 + ......+ 1/2006.1004
CMR: A/B thuộc Z ( số nguyên )
cho A = \(\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
CMR \(\frac{7}{12}< A< \frac{5}{6}\)
cho A= \(\frac{1}{1.2^2}+\frac{1}{2.3^2}+\frac{1}{3.4^2}+...+\frac{1}{49.50^2}\)
B= \(\frac{1}{2^{ }}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{50^2}\)
Chứng minh : A < \(\frac{1}{2}\)<B
chứng minh rằng:
\(\frac{1.2-1}{2!}+\frac{2.3-1}{3!}+\frac{3.4-1}{4!}+...+\frac{99.100-1}{100!}< 2\)
mình ngu toán chúng minh (hép mi)
So sánh A và B với \(\dfrac{1}{2}\) biết :
\(A=\dfrac{1}{1.2^2}+\dfrac{1}{2.3^2}+\dfrac{1}{3.4^2}+........+\dfrac{1}{49.50^2}\) và
\(B=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+.......+\dfrac{1}{50^2}\)
