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31 tháng 3 2021
Ta có: \(A=\dfrac{1+\dfrac{1}{3}+\dfrac{1}{5}+\dfrac{1}{7}+...+\dfrac{1}{97}+\dfrac{1}{99}}{\dfrac{1}{1\cdot99}+\dfrac{1}{3\cdot97}+\dfrac{1}{5\cdot95}+...+\dfrac{1}{97\cdot3}+\dfrac{1}{99\cdot1}}\)
\(\Leftrightarrow\dfrac{A}{100}=\dfrac{1+\dfrac{1}{3}+\dfrac{1}{5}+\dfrac{1}{7}+...+\dfrac{1}{97}+\dfrac{1}{99}}{\dfrac{100}{1\cdot99}+\dfrac{100}{3\cdot97}+\dfrac{100}{5\cdot95}+...+\dfrac{100}{97\cdot3}+\dfrac{100}{99\cdot1}}\)
\(\Leftrightarrow\dfrac{A}{100}=\dfrac{1+\dfrac{1}{3}+\dfrac{1}{5}+\dfrac{1}{7}+...+\dfrac{1}{97}+\dfrac{1}{99}}{1+\dfrac{1}{99}+\dfrac{1}{3}+\dfrac{1}{97}+\dfrac{1}{5}+\dfrac{1}{95}+...+\dfrac{1}{97}+\dfrac{1}{3}+\dfrac{1}{99}+1}\)
\(\Leftrightarrow\dfrac{A}{100}=\dfrac{1+\dfrac{1}{3}+\dfrac{1}{5}+\dfrac{1}{7}+...+\dfrac{1}{97}+\dfrac{1}{99}}{2\left(1+\dfrac{1}{3}+\dfrac{1}{5}+\dfrac{1}{7}+...+\dfrac{1}{97}+\dfrac{1}{99}\right)}\)
\(\Leftrightarrow\dfrac{A}{100}=\dfrac{1}{2}\)
hay A=50
KL
26 tháng 7 2017
a, \(4\times\left(-\dfrac{1}{2}\right)^3-2\times\left(-\dfrac{1}{2}\right)^2+3\times\left(-\dfrac{1}{2}\right)+1\)
\(=\left(-\dfrac{1}{2}\right)\left[\left(4\times-\dfrac{1}{2}\right)-\left(2\times-\dfrac{1}{2}\right)+3\right]+1\)
\(=\left(-\dfrac{1}{2}\right)\left(-2+1+3\right)+1\)
\(=\left(-\dfrac{1}{2}\right)2+1\)
\(=-1+1\)
\(=0\)
@Trịnh Thị Thảo Nhi
29 tháng 4 2018
a, 4×(−12)3−2×(−12)2+3×(−12)+14×(−12)3−2×(−12)2+3×(−12)+1
=(−12)[(4×−12)−(2×−12)+3]+1=(−12)[(4×−12)−(2×−12)+3]+1
=(−12)(−2+1+3)+1=(−12)(−2+1+3)+1
=(−12)2+1=(−12)2+1
=−1+1=−1+1
=0=0
19 tháng 1 2019
Đặt A = \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}\)
=> A = \(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
=> A = 1 - \(\dfrac{1}{100}\) = \(\dfrac{99}{100}\)
=> 1 = \(\dfrac{100}{100}\)
=> A < 1
LT
18 tháng 6 2019
A = 11.2+12.3+13.4+...+199.10011.2+12.3+13.4+...+199.100
=> A = 1−12+12−13+13−14+...+199−11001−12+12−13+13−14+...+199−1100
=> A = 1 - 11001100 = 9910099100
=> 1 = 100100100100
=> A < 1
K
29 tháng 3 2018
\(\dfrac{2}{1.2}+\dfrac{2}{2.3}+\dfrac{2}{3.4}+...+\dfrac{2}{99.100}\)
\(=2.\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}\right)\)
\(=2.\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)\)
\(=2.\left(1-\dfrac{1}{100}\right)=2.\dfrac{99}{100}=\dfrac{99}{50}\)
31 tháng 12 2022
a: =>x*2/15=2/7
=>x=2/7:2/15=2/7*15/2=15/7
b: x=3:7/5=15/7
c: x=-1/2:4/9=-1/2*9/4=-9/8
d: x=-8/3:3/8=-64/9
g: =>4/11x=2/5+1/3=6/15+5/15=11/15
=>x=11/15:4/11=121/60
l: =>1/4:x=1-3/2=-1/2
=>x=-1/4:1/2=-1/4*2=-1/2
k: =>x:7=-1/3+5/2=-2/6+15/6=13/6
=>x=91/6
17 tháng 10 2017
\(E=\dfrac{11.3^{29}-3^{2^{15}}}{2.3^{14}.2.3^{14}}\)
\(=\dfrac{11.3-3^{30}}{2^2}=\dfrac{33-3^{30}}{4}\)
MV
27 tháng 4 2017
\(\dfrac{200-\left(3+\dfrac{2}{3}+\dfrac{2}{4}+\dfrac{2}{5}+...+\dfrac{2}{100}\right)}{\dfrac{1}{2}+\dfrac{2}{3}+\dfrac{3}{4}+...+\dfrac{99}{100}}\\ =\dfrac{200-2-\left(1+\dfrac{2}{3}+\dfrac{2}{4}+...+\dfrac{2}{100}\right)}{\left(1-\dfrac{1}{2}\right)+\left(1-\dfrac{1}{3}\right)+\left(1-\dfrac{1}{4}\right)+...+\left(1-\dfrac{99}{100}\right)}\\ =\dfrac{198-\left(\dfrac{2}{2}+\dfrac{2}{3}+\dfrac{2}{4}...+\dfrac{2}{100}\right)}{\left(1+1+1+...+1\right)-\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{100}\right)}\\ =\dfrac{2\cdot99-2\cdot\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{100}\right)}{99-\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{100}\right)}\\ =\dfrac{2\cdot\left[99-\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{100}\right)\right]}{99-\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{100}\right)}=2\left(đpcm\right)\)
21 tháng 5 2022
a: \(=\left(\dfrac{17}{10}+\dfrac{70}{10}-\dfrac{87}{10}\right):\left(\dfrac{23}{4}-\dfrac{11}{4}+\dfrac{9}{25}\right)\cdot\left(12,98\cdot0,25\right)+12,5\)
\(=0:\left(3+\dfrac{9}{25}\right)\cdot\left(12,98+0,25\right)+12,5\)
=12,5
b: \(=\dfrac{13}{12}\cdot\dfrac{27}{5}\cdot2\cdot\dfrac{34}{9}\cdot2\cdot\dfrac{2}{17}\)
\(=\dfrac{13}{12}\cdot2\cdot\dfrac{27}{5}\cdot\dfrac{34}{9}\cdot\dfrac{4}{17}\)
\(=\dfrac{13}{6}\cdot\dfrac{27}{5}\cdot\dfrac{8}{9}=\dfrac{8}{6}\cdot3\cdot\dfrac{13}{5}=4\cdot\dfrac{13}{5}=\dfrac{52}{5}\)
`1/3xx3+3:2/3`
`=3/3+3xx3/2`
`=1+9/2`
`=2/2+9/2`
`=11/2`