1/1x2+3/2x5+5/5x10+4/10x14+6/14x20
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6 x5/3=10 7:8/4+1/3=23/6 2x5/8-1/2=3/4 3/7:5/6x1/4=9/70
Ta có A = \(\frac{1}{1.2}+\frac{3}{2.5}+\frac{9}{5.14}+\frac{23}{14.37}+\frac{15}{37.52}+\frac{1967}{52.2019}\)(sửa lại đề)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{14}+\frac{1}{14}-\frac{1}{37}+\frac{1}{37}-\frac{1}{52}+\frac{1}{52}-\frac{1}{2019}\)
\(=1-\frac{1}{2019}=\frac{2018}{2019}< 1\)
=> A < 1 (ĐPCM)
`1)1/2:2/3 .... 2/3 : 1/2`
`=>1/2xx3/2 .... 2/3xx2`
`=>3/4 .... 4/3`
Vì `3/4 < 1` và `4/3>1`
`=>3/4<4/3`
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`4/7:2/5 ... 4/7 : 3/5`
`=>4/7xx5/2....4/7xx5/3`
`=>20/14...20/21`
`=>10/7...20/21`
Vì `10/7>1` và `20/21<1`
`=>10/7>20/21`
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`4/15:4/7....2/5xx10/3`
`=>4/15xx7/4...20/15`
`=>7/15...20/15`
Vì `7<20` nên `7/15<20/15`
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`5/6...15/18-11/18`
`=>5/6...4/18`
Ta có : MSC : `18`
`5/6 = 15/18`
Vì `15>4` nên `5/6 > 4/18`
\(C=1.2+2.3+3.4+...+n\left(n+1\right)\\ \Rightarrow3.C=1.2.3+2.3.3+3.4.3+..+n\left(n+1\right).3\\ \Rightarrow3.C=1.2.3+2.3.4-1.2.3+....+n\left(n+1\right)\left(n+2\right)-\left(n-1.n.\left(n+1\right)\right)\\ \Rightarrow3.C=n\left(n+1\right)\left(n+2\right)\\ \Rightarrow C=\dfrac{n\left(n+1\right)\left(n+2\right)}{3}\)
Cái D tính TT
(2 ^ 2 + 1) x (2 ^ 2 - 1) x (2 ^ 4 - 4 ^ 2)
= (2 ^ 2 + 1) x (2 ^ 2 - 1) x ( 16 - 16 )
= (2 ^ 2 + 1) x (2 ^ 2 - 1) x 0
= 0
( 7 ^ 3 + 7 ^ 5) x (5 ^ 4 + 5 ^ 6) x ( 3 ^ 3 x 3 - 9 ^ 2)
= ( 7 ^ 3 + 7 ^ 5) x (5 ^ 4 + 5 ^ 6) x ( 3 ^ 4 - 3 ^ 4)
= ( 7 ^ 3 + 7 ^ 5) x (5 ^ 4 + 5 ^ 6) x 0
= 0
2 ^ 2 x 5 [( 5 ^ 2 + 2 ^ 3) : 11] - 2 ^ 4 + 2 x 10 ^ 3
= 4 x 5 [( 25 + 8 ) : 11] - 8 + 2 x 1000
= 20 ( 33 : 11) - 8 + 2000
= 20 x 3 - 8 + 2000
= 60 - 8 + 2000
= 2052
A = \(\dfrac{1}{1\times2}\) + \(\dfrac{3}{2\times5}\) + \(\dfrac{5}{5\times10}\) + \(\dfrac{4}{10\times14}\) + \(\dfrac{6}{14\times20}\)
A = \(\dfrac{1}{1}\) - \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) - \(\dfrac{1}{5}\) + \(\dfrac{1}{5}\) - \(\dfrac{1}{10}\) + \(\dfrac{1}{10}\) - \(\dfrac{1}{14}\) + \(\dfrac{1}{14}\) - \(\dfrac{1}{20}\)
A = \(\dfrac{1}{1}\) - \(\dfrac{1}{20}\)
A = \(\dfrac{19}{20}\)