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TM
4
KB
28 tháng 4 2015
A = 5/1.2 + 5/2.3 +...+ 5/99.100
2A = 10/1.2 + 10/2.3 +...+ 10/99.100
2 A = 5/1-5/2+5/2-5/3+5/3-5/4+...+5/99-5/100
2A=5/1-5/100
2A=9/2 => A=9/2:2=9/4
cho 1 đ-ú-n-g nha bạn!!
28 tháng 4 2015
99/20 đảm bảo
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C
1
3 tháng 5 2016
\(A=5\left(\frac{1}{1.2}+\frac{1}{2.3}+.........+\frac{1}{99.100}\right)\)
\(=5\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+............+\frac{1}{99}-\frac{1}{100}\right)\)
\(=5\left(1-\frac{1}{100}\right)\)
\(=5.\frac{99}{100}\)
\(=\frac{99}{20}\)
T
3
MX
18 tháng 4 2017
A =\(\frac{5}{1.2}+\frac{5}{2.3}+\frac{5}{3\cdot4}+...+\frac{5}{99.100}\)
A = 5 x (\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\) )
A = 5 x \(\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\right)\)
A = 5 x \(\left(1-\frac{1}{100}\right)\)
A = 5 x \(\frac{99}{100}\)
A = \(\frac{495}{100}\)
A= \(\frac{99}{20}\)
18 tháng 4 2017
Ta co : A =5.(1/1.2+1/2.3+1/3.4+....+1/99.100)
A= 5.(1-1/2+1/2-1/3+1/3-1/4+.....+1/99-1/100)
Rut gon tung so ta co :A=5.(1-1/100)
A=5.99/100
A=1.99/50=99/50
DT
1
K
27 tháng 6 2015
\(\frac{5}{1.2}+\frac{5}{2.3}+...+\frac{5}{99.100}-2x=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{97.99}\)
\(5\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\right)-2x=\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{97.99}\right)\)
\(5\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\right)-2x=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\right)\)\(5\left(1-\frac{1}{100}\right)-2x=\frac{1}{2}\left(1-\frac{1}{99}\right)\)
\(5.\frac{99}{100}-2x=\frac{1}{2}.\frac{98}{99}\)
\(\frac{99}{20}-2x=\frac{49}{99}\)
\(2x=\frac{99}{20}-\frac{49}{99}\)
\(2x=\frac{8821}{1980}\)
\(x=\frac{8821}{1980}:2\)
\(x=\frac{8821}{3960}\)
V
1
NN
22 tháng 3 2023
\(=5\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}\right)\)
\(=5.\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)\)
\(=5\left(1-\dfrac{1}{100}\right)\)
\(=5.\dfrac{99}{100}=\dfrac{99}{20}\)
DT
2
24 tháng 6 2015
Cai phan 1+3+5+...+99 chac em biet lam roi phai ko? Con 3/1.2+3/2.3+3/3.4+...+3/99.100 thi em cu tach lam sao cho tro thanh dang ban dau thi lam . Anh phai nghi roi !~ Neu chieu anh ranh ranh thi len giai tiep . BYE BYE
K
24 tháng 6 2015
\(\frac{3}{1.2}+\frac{3}{2.3}+\frac{3}{3.4}+...+\frac{3}{99.100}+4x=1+3+5+...+99\)
\(3\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\right)+4x=\left(1+99\right)+\left(3+97\right)+\left(5+95\right)+...+\left(49+51\right)\)\(3\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\right)+4x=100+100+100+...+100\)\(3\left(1-\frac{1}{100}\right)+4x=100.25\)
\(3.\frac{99}{100}+4x=2500\)
\(\frac{297}{100}+4x=2500\)
\(4x=2500-\frac{297}{100}\)
\(4x=2500-2,97\)
\(4x=2497,03\)
\(x=624,2575\)
\(x=2497,03:4\)
KN
2
24 tháng 3 2019
1/5 .B=1/1×2+1/2×3+.....+1/99×100
1/5 .B=2-1/1×2+3-2/2×3+......+100-99/99×100
1/5 .B=1/1-1/2+1/2-1/3+......+1/99-1/100
1/5.B=1-1/100
1/5. B=99/100
B=99/100÷1/5
B=99/20
24 tháng 3 2019
\(B=\frac{5}{1.2}+\frac{5}{2.3}+.....+\frac{5}{99.100}\)
\(B=5-\frac{5}{2}+\frac{5}{2}-\frac{5}{3}+.....+\frac{5}{99}-\frac{5}{100}\)
\(B=5-\frac{5}{100}\)
\(B=\frac{500}{100}-\frac{5}{100}\)
\(B=\frac{495}{100}=\frac{99}{20}\)
NV
cho bieu thuc A=\(\frac{5}{1.2}+\frac{5}{2.3}+...+\frac{5}{99.100}\)
hay chung to rang 5>A
giup mk nha
2
HP
24 tháng 2 2016
\(A=\frac{5}{1.2}+\frac{5}{2.3}+...+\frac{5}{99.100}\)
\(A=\frac{5}{1}-\frac{5}{2}+\frac{5}{2}-\frac{5}{3}+...+\frac{5}{99}-\frac{5}{100}=\frac{5}{1}-\frac{5}{100}=\frac{99}{20}<\frac{100.}{20}=5\)
=>A<5 hay 5>A(đpcm)
24 tháng 2 2016
Ta có
\(A=\frac{5}{1.2}+\frac{5}{2.3}+...+\frac{5}{99.100}=5\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\right)\)
\(=5\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\right)=5.\left(1-\frac{1}{100}\right)\)
Vì \(1-\frac{1}{100}<1\)
=>\(5\left(1-\frac{1}{100}\right)<5.1=5\)
Vậy A<5
MH
25 tháng 2 2016
A = \(5.\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\right)\)
= \(5.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\right)\)
= \(5.\left(1-\frac{1}{100}\right)\)
= \(5-\frac{5}{100}<5\)
=> A < 5 (Đpcm).
A = 5 .(1/1.2 +1/2.3 + 1/3.4 +.....1/99.100)
A= 5 . (1/1 - 1/2 +1/2 -1/3 +1/3 -1/4 ........ +1/99 - 1/100)
A= 5 . (1/1 - 1/100)
A= 5.99/100
A= 99/20
A=5.(1-1/2+1/2-1/3 + .... + 1/99 - 1/100)
= 5.(1-1/100)
=5 . 99/100
= 99/20