tính nhanh:
a,101^2
b,9999^2
c,47.53
d,991.1009
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31 tháng 8 2021
Câu 3:
a: \(49^2=2401\)
b: \(51^2=2601\)
c: \(99\cdot100=9900\)
BA
13 tháng 1 2016
A = 1/15 + 1/35 + 1/63 + 1/99 +........1/9999
A = 1/3x5 + 1/5x7 + 1/7x9 + .......+ 1/99x101
A = 1/2 x (1/3 - 1/5 + 1/5 - 1/7 + 1/7 - 1/9 +..... + 1/99 - 1/101)
A = 1/2 X ( 1/3 - 1/99)
A = 1/2 x ( 1/3 - 1/101)
A = 1/2 ( 98/303)
A = 49/303
LT
2
H9
HT.Phong (9A5)
CTVHS
13 tháng 8 2023
a) \(\dfrac{2}{5}+\dfrac{4}{5}\times\dfrac{5}{2}\)
\(=\dfrac{2}{5}+\dfrac{4\times5}{5\times2}\)
\(=\dfrac{2}{5}+\dfrac{4}{2}\)
\(=\dfrac{2}{5}+2\)
\(=\dfrac{2}{5}+\dfrac{10}{5}\)
\(=\dfrac{12}{5}\)
b) \(\dfrac{2008}{2009}-\dfrac{2009}{2008}+\dfrac{1}{2009}+\dfrac{2007}{2008}\)
\(=\left(1-\dfrac{1}{2009}\right)-\left(1+\dfrac{1}{2008}\right)+\dfrac{1}{2009}+\left(1-\dfrac{1}{2008}\right)\)
\(=1-\dfrac{1}{2009}-1-\dfrac{1}{2008}+\dfrac{1}{2009}+1-\dfrac{1}{2008}\)
\(=\left(1-1+1\right)-\left(\dfrac{1}{2009}-\dfrac{1}{2009}\right)-\left(\dfrac{1}{2008}+\dfrac{1}{2008}\right)\)
\(=1-\dfrac{2}{2008}\)
\(=\dfrac{2008}{2008}-\dfrac{2}{2008}\)
\(=\dfrac{2006}{2008}\)
\(=\dfrac{1003}{1004}\)
13 tháng 8 2023
a: =2/5+4/2
=2/5+2
=12/5
b: \(=1-\dfrac{1}{2009}-1-\dfrac{1}{2008}+\dfrac{1}{2009}+1-\dfrac{1}{2008}\)
\(=1-\dfrac{2}{2008}=1-\dfrac{1}{1004}=\dfrac{1003}{1004}\)
NC
1
3 tháng 2 2016
A = 1/15 + 1/35 + 1/ 63 + 1/99 + ...+ 1/9999
A = 1/(3x5) + 1/(5x7) + 1/(7x9) + 1/(9x11) + ... + 1/(99 x 101)
Ax2 = 2/(3x5) + 2/(5x7) + 2/(7x9) + 2/(9x11) + ... + 2/(99 x 101)
Ax2 = 1/3 – 1/5 + 1/5 – 1/7 + 1/7 – 1/9 + 1/9 – 1/11 + ...+ 1/99 – 1/101
Ax2 = 1/3 – 1/101 = 98/303
A = 98/303 : 2
A = 49/303
tớ không chắc nhé
28 tháng 6 2021
`a)1001^2`
`=(1000+1)^2=1000000+2000+1`
`=1002001`
`b)29,9.30,1`
`=(30-0,1)(30+0,1)`
`=30^2-0,1^2`
`=900-0,01=899,99`
`c)199^2=(200-1)^2`
`=40000-400+1`
`=39601`
`d)84^2-16^2`
`=(84-16)(84+16)`
`=100.68`
`=6800`
`e)313^2-312^2`
`=(313-312)(313+312)`
`=625`
`f)47.53`
`=(50-3)(50+3)`
`=2500-9=2491`
\(101^2=\left(100+1\right)^2=10000+200+1=10201\\ 9999^2=\left(10000-1\right)^2=100000000-20000+1=99980001\\ 47\cdot53=\left(50-3\right)\left(50+3\right)=2500-9=2491\\ 991\cdot1009=\left(1000-9\right)\left(1000+9\right)=1000000-81=999919\)
a: \(101^2=10201\)
b: \(9999^2=99980001\)
c: \(47\cdot53=2491\)
d: \(991\cdot1009=999919\)