Cho A=1/51+1/52+1/53+...+1/100 và B=1/2+1/12+1/30+...+1/9900.
Tính A-B
Cho A=1/51+1/52+1/53+...+1/100
Cho B=1/2+1/12+1/30+...+1/9900
Tính A-B
a, Tìm tất cả các cặp số nguyên (x,y) thỏa mãn: x/2- 2/y=1/2
b,Tinh hợp lí: A=(1/2+1/12+1/30+...+1/9900) : (-6/51- 6/52- 6/53-...- 6/100)
a=(\(\dfrac{1}{2}+\dfrac{1}{12}+\dfrac{1}{30}+...+\dfrac{1}{9900}\)):(\(\dfrac{-6}{51}-\dfrac{6}{52}-\dfrac{6}{53}-...-\dfrac{6}{100}\))
giúp mik giải nhé
cảm ơn !
tìm B=(1/2+1/6+1/12+...+1/9900)/(1/51+1/52+1/53+...+1/100)
1/2+1/12+1/30+...+1/9120+1/9506+1/9900. / 50-50/51-51/52-...-97/98-98/99-99/100
cho A = 1/1*2+1/3*4+...+1/99*100 và B= 2015/51+2015/52+2015/53+...+2015/100. Chứng minh rằng B chia hết cho A
cho A = 1/1*2+1/3*4+...+1/99*100 và B= 2015/51+2015/52+2015/53+...+2015/100. Chứng minh rằng B chia hết cho A
Ta có : \(A=\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(=\left(1+\frac{1}{3}+...+\frac{1}{99}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{99}+\frac{1}{100}\right)-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{99}+\frac{1}{100}\right)-\left(1+\frac{1}{2}+...+\frac{1}{50}\right)\)
\(=\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}\)
\(B=\frac{2015}{51}+\frac{2015}{52}+...+\frac{2015}{100}\)
\(=2015\left(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}\right)\)
\(\Rightarrow\) \(\frac{B}{A}=\frac{2015\left(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}\right)}{\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}}=2015\)
\(\Rightarrow\) \(B⋮A\)
cho A = 1/1*2+1/3*4+...+1/99*100 và B= 2015/51+2015/52+2015/53+...+2015/100. Chứng minh rằng B chia hết cho A
Cho A = 1/1.2+1/3.4+1/5.6+...+1/99.100
a Chứng minh A= 1/51+1/52+1/53+...+1/100
b Chứng minh 7/12<A<5/6
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