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I
6 tháng 6 2021
1
Ta có :A=1/1.2+1/3.4+...+1/99.100=1/2+1/12+...+1/9900
7/12=1/2+1/12
Vì 1/2+1/12<1/2+1/12+...+1/9900
Nên: 7/12<A (1)
Lại có:A=1/1.2+1/3.4+...+1/99.100
=1-1/2+1/3-1/4+...+1/99-1/100
=(1-1/2+1/3)+(-1/4+1/5-1/6)+...+(-1/98+1/99-1/100)
5/6=1-1/2+1/3
vì: 1-1/2+1/3 < (1-1/2+1/3)+(-1/4+1/5-1/6)+...+(-1/98+1/99-1/100)
nên 5/6 < A (2)
Từ (1) và (2) suy ra 7/12<A<5/6
3 tháng 9 2016
a)A = 1 / (1*2) + 1 / (3*4) + ... + 1 / (99*100) > 1 / (1*2) + 1 / (3*4) = 1 / 2 + 1 / 12 = 7 / 12 ♦
A = 1 / (1*2) + 1 / (3*4) + ... + 1 / (99*100) = (1 - 1 / 2) + (1 / 3 - 1 / 4) + ... + (1 / 99 - 100) =
(1 - 1 / 2 + 1 / 3) - (1 / 4 - 1 / 5) - (1 / 6 - 1 / 7) - ... - (1 / 98 - 1 / 99) - 1 / 100 <
1 - 1 / 2 + 1 / 3 = 5 / 6 ♥
♦, ♥ => 7 / 12 < A < 5 / 6
b)ta có:
1/1.2+1/3.4+1/5.6+...+1/49.50
=>1-1/2+1/3-1/4+1/5-1/6+...+1/49-1/50
=>(1+1/3+1/5+1/7+...+1/49)-(1/2+1/4+1/6+...+1/50)
=>(1+1/2+1/3+...+1/49+1/50)-(1/2+1/4+1/6+...+1/50).2
=>(1+1/2+1/3+...+1/49+1/50) -( 1+1/2+1/3+...+1/25)
=>1/26+1/27+1/28+...+1/50=1/26+1/27+1/28+...+1/50
hay 1/1.2+1/3.4+1/5.6+...+1/49.50=1/26+1/27+1/28+...+1/50
NV
Nguyễn Việt Lâm
Giáo viên
24 tháng 6 2019
\(\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{49.50}=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{60}\)
\(=1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{49}-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{50}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{50}\right)-\left(\frac{1}{2}+\frac{1}{2}+\frac{1}{4}+\frac{1}{4}+...+\frac{1}{50}+\frac{1}{50}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{50}\right)-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{50}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{50}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{25}\right)\)
\(=\frac{1}{26}+\frac{1}{27}+...+\frac{1}{50}\)
2/ \(A=\frac{1}{2}+\frac{1}{12}+\frac{1}{5.6}+...+\frac{1}{99.100}\)
\(A=\frac{7}{12}+\frac{1}{5.6}+\frac{1}{7.8}+...+\frac{1}{99.100}>\frac{7}{12}\)
Tương tự câu trên ta có: \(A=\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}\)
\(A=\frac{1}{51}+...+\frac{1}{60}+\frac{1}{61}+...+\frac{1}{70}+\frac{1}{71}+...+\frac{1}{80}+\frac{1}{81}+...+\frac{1}{90}+\frac{1}{91}+...+\frac{1}{100}\)
\(A< \frac{1}{50}+...+\frac{1}{50}+\frac{1}{60}+...+\frac{1}{60}+\frac{1}{70}+...+\frac{1}{70}+\frac{1}{80}+...+\frac{1}{80}+\frac{1}{90}+...+\frac{1}{90}\)
\(A< 10.\frac{1}{50}+10.\frac{1}{60}+10.\frac{1}{70}+10.\frac{1}{80}+10.\frac{1}{90}\)
\(A< \frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+\frac{1}{9}< \frac{5}{6}\)
12 tháng 7 2017
2, a-b=ab => a=ab+b => a=b(a+1)
thay a=b(a+1) vào a:b ta có: => b:b(a+1)=a+1
Theo bài ra ta có: a:b=a-b
=> a+1=a-b
=>-b=1
=> b=-1
Thay b=-1 vào a-b=ab ta có : a-(-1)=-a
=> a +1=-a
=>a=-1/2
Vậy a=-1/2. b=-1
S
26 tháng 6 2019
\(\frac{1}{1.2}+\frac{1}{3.4}+......+\frac{1}{49.50}=1-\frac{1}{2}+\frac{1}{3}-....+\frac{1}{49}-\frac{1}{50}=\left(1+\frac{1}{3}+....+\frac{1}{49}\right)-\left(\frac{1}{2}+\frac{1}{4}+.....+\frac{1}{50}\right)=\left(1+\frac{1}{2}+.....+\frac{1}{50}\right)-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{50}\right)=\left(1+\frac{1}{2}+....+\frac{1}{50}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+....+\frac{1}{25}\right)=\frac{1}{26}+\frac{1}{27}+....+\frac{1}{50}\left(đpcm\right)\)
\(theocaua\Rightarrow A=\frac{1}{26}+\frac{1}{27}+......+\frac{1}{50}>\frac{1}{30}+\frac{1}{30}+...+\frac{1}{30}\left(5sohang\right)+\frac{1}{40}+\frac{1}{40}+....+\frac{1}{40}\left(10sohang\right)+\frac{1}{50}+\frac{1}{50}+....+\frac{1}{50}\left(10sohang\right)=\frac{1}{4}+\frac{1}{5}+\frac{1}{6}=\frac{37}{60}>\frac{35}{60}=\frac{7}{12}\left(1\right)\)
\(A=\frac{1}{26}+\frac{1}{27}+....+\frac{1}{50}< \frac{1}{25}+\frac{1}{25}+...+\frac{1}{25}\left(5sohang\right)+\frac{1}{30}+\frac{1}{30}+....+\frac{1}{30}\left(10sohang\right)+\frac{1}{40}+\frac{1}{40}+.....+\frac{1}{40}\left(10sohang\right)=\frac{1}{4}+\frac{1}{3}+\frac{1}{5}=\frac{47}{60}< \frac{5}{6}=\frac{50}{60}\left(2\right)\) \(\left(1\right);\left(2\right)\Rightarrow\frac{7}{12}< A< \frac{5}{6}\)
21 tháng 10 2016
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)
=\(\left(\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{50}\right)-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{50}\right)\)
=\(\left(\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{50}\right)-\left(\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{25}\right)\)
=\(\frac{1}{26}+\frac{1}{27}+\frac{1}{28}+...+\frac{1}{50}\)
7 tháng 9 2015
\(A=\frac{1}{1}-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{99}-\frac{1}{100}=\left(\frac{1}{1}+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{99}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)\)
\(A=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{99}\right)+\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)-2.\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)\)
\(A=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{50}\right)=\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}\)
+) \(A=\left(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{75}\right)+\left(\frac{1}{76}+...+\frac{1}{100}\right)>\left(\frac{1}{75}+...+\frac{1}{75}\right)+\left(\frac{1}{100}+...+\frac{1}{100}\right)\)
=> \(A>\frac{25}{75}+\frac{25}{100}=\frac{1}{3}+\frac{1}{4}=\frac{7}{12}\)
+) \(A=\left(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{75}\right)+\left(\frac{1}{76}+...+\frac{1}{100}\right)
