cho A= 1/1.2 + 1/3.4 + 1/5.6 + ... + 1/99.100
a, chứng tỏ : A= 1/51 + 1/52 + 1/53 + ... + 1/99.100
b, chứng tỏ 7/12< A< 5/6
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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Cho: A = 1/1.2 + 1/3.4 + 1/5.6 + .... +1/99.100
chứng tỏ rằng: 7/12 < A < 5/6
\(A=\frac{1}{2}+\frac{1}{12}+...+\frac{1}{9900}>\frac{1}{2}+\frac{1}{12}=\frac{7}{12}\)
\(A=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{99}-\frac{1}{100}=\left(1-\frac{1}{2}+\frac{1}{3}\right)-\left(\frac{1}{4}-\frac{1}{5}\right)-...-\left(\frac{1}{98}-\frac{1}{99}\right)-\frac{1}{100}
Ta có: A=1/1.2+1/3.4+1/5.6+...+1/99.100
=1-1/2+1/3-1/4+1/5-1/6+...+1/99-1/100
=1+1/2+1/3+1/4+1/5+1/6+...+1/99+1/100-2(1/2+1/4+1/6+...+1/100)
=1+1/2+1/3+1/4+1/5+1/6+...+1/99+1/100-(1+1/2+1/3+1/4+...+1/50)
=1/26+1/27+1/28+...+1/100)
Do đó A=(1/51+1/52+...+1/75)+(1/76+1/77+...+1/100)
Ta có 1/51>1/52>...>1/75 và 1/76>1/77>...>1/100 nên
A>1/75.25+1/100.25=1/3+1/4=7/12
A<1/51.25+1/76.25<1/50.25+1/75.25=1/2+1/3=5/6
Vậy nên 7/12<A<5/6
Cho A=1/1.2+1/3.4+1/5.6+...+1/99.100. Chứng tỏ rằng 17/12 < A < 5/6
so sánh A và B . biết A= 1/1.2 + 1/3.4 + 1/5.6 + ......+ 1 / 99.100
B = 2021/ 51 + 2021/52 + 2021/53 + .... + 2021/100
so sánh A và B . biết A= 1/1.2 + 1/3.4 + 1/5.6 + ......+ 1 / 99.100
B = 2021/ 51 + 2021/52 + 2021/53 + .... + 2021/100
=> A < B
chắc vại-.-
tui hok giỏi toán lém
Cho \(A=\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+....\frac{1}{99.100}.\)Chứng minh rằng:
a.\(A=\frac{1}{51}+\frac{1}{52}+\frac{1}{53}+....+\frac{1}{100}.\)
b.\(\frac{7}{12}< A< \frac{5}{6}.\)
A=1/1.2+1/3.4+1/5.6+...+1/99.100 = ? B=2015/51+2015/52+2015/53+...+2015/100
Chứng minh rằng B/A là 1 số nguyên
A=1/1.2+1/3.4+1/5.6+...+1/99.100 = ? B=2015/51+2015/52+2015/53+...+2015/100
Chứng minh rằng B/A là 1 số nguyên
Bài 1: Cho A=\(\dfrac{1}{1.2}+\dfrac{1}{3.4}+\dfrac{1}{5.6}+...+\dfrac{1}{99.100}\)
a) Chứng minh: A=\(\dfrac{1}{51}+\dfrac{1}{52}+\dfrac{1}{53}+...+\dfrac{1}{100}\)
b) Chứng minh: A<\(\dfrac{5}{6}\)
a) $A=\dfrac{1}{1.2}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}$
$=>A=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}$
$=>A=(1+\dfrac{1}{3}+...+\dfrac{1}{99})-(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{100})$
$=>A=(1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{99}+\dfrac{1}{100})-(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{100}.2)$
$=>A=(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{100})-(1+\dfrac{1}{2}+...+\dfrac{1}{50})$
$=>A=\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{100}$
b) Ta có : $A=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}$
$=>A=(1-\dfrac{1}{2}+\dfrac{1}{3})-(\dfrac{1}{4}-\dfrac{1}{5})-...-(\dfrac{1}{98}-\dfrac{1}{99})-\dfrac{1}{100}$
$=>A<1-\dfrac{1}{2}+\dfrac{1}{3}=\dfrac{5}{6}$