A=1+32+34+...+32008
Trả lời giúp mình với.
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Rút gọn: (3 + 1)(3^2 + 1)(3^4 + 1)(3^8 + 1)(3^16 + 1)(3^32 + 1)
A=(3 + 1)(3^2 + 1)(3^4 + 1)(3^8 + 1)(3^16 + 1)(3^32 + 1)
2A=2(3 + 1)(3^2 + 1)(3^4 + 1)(3^8 + 1)(3^16 + 1)(3^32 + 1)
2A=(3-1)(3 + 1)(3^2 + 1)(3^4 + 1)(3^8 + 1)(3^16 + 1)(3^32 + 1)
2A=(3^2-1)(3^2 + 1)(3^4 + 1)(3^8 + 1)(3^16 + 1)(3^32 + 1)
2A=(3^4-1)(3^4 + 1)(3^8 + 1)(3^16 + 1)(3^32 + 1)
2A=(3^8-1)(3^8 + 1)(3^16 + 1)(3^32 + 1)
2A=(3^16-1)(3^16 + 1)(3^32 + 1)
2A=(3^32 - 1)(3^32 + 1)
2A=3^64-1
=>A=(3^64-1) /2
ta có: \(\frac{31+32+35}{34}=\frac{31}{34}+\frac{32}{34}+\frac{35}{34}.\)
mà \(\frac{31}{32}>\frac{31}{34};\frac{32}{33}>\frac{32}{34}\)
\(\Rightarrow\frac{31}{32}+\frac{32}{33}+\frac{35}{34}>\frac{31}{34}+\frac{32}{34}+\frac{35}{34}=\frac{31+32+35}{34}\)
1. So sánh bằng cách nhanh nhất:
a. \(\frac{28}{31}>\frac{28}{33}\)
b.\(\frac{28}{81}>\frac{11}{34}\)
c. \(\frac{13}{11}>\frac{31}{32}\)
d. \(\frac{1994}{1995}>\frac{36}{37}\)
Giúp mình với!!!
a,(-19)+(-32)+2019+52 = -19 + 2019 + 52 - 32
= 2000 + 20
= 2020
b,34-(15+34-40)= 34 - 15 -34 + 40
= -15 + 40
= 25
c,-11.(-4)+6.(-5)= 44 + ( -30)
= 14
d,88.(12-67)-12.(88+67)= 88.12 - 5896 - 12.88 + 804
= -5896 + 804
= -5092
hok tốt!
a,(-19)+(-32)+2019+52
=(-19+2019)+(-32+52)
=2000+20
=2020
b,34-(15+34-40)
=34-15-34+40
=(34-34)-15+40
=0-15+40
=-15+40
=-25
c,-11.(-4)+6.(-5)
=44+(-30)
=14
d,88.(12-67)-12.(88+67)
=88.(-55)-12.155
=-4840-1860
=6700
`A=4(3^2+1)(3^4+1)...(3^64+1)`
`=>2A=(3^2-1)(3^2+1)(3^4+1)...(3^64+1)`
- Ta có:
`(3^2-1)(3^2+1)=3^4-1`
`(3^4-1)(3^4+1)=3^16-1`
`....`
`(3^64-1)(3^64+1)=3^128-1`
Suy ra `2A=3^128-1=B`
`=>A<B`
a, \(\dfrac{90}{37}-\dfrac{38}{25}-\dfrac{8}{25}-\dfrac{4}{25}\)
= \(\dfrac{90}{37}\) - \(\dfrac{38+8+4}{25}\)
= \(\dfrac{90}{37}\) - 2
= \(\dfrac{16}{37}\)
\(\dfrac{24}{29}\) + \(\dfrac{32}{41}\) + \(\dfrac{34}{29}\) + \(\dfrac{50}{41}\)
=(\(\dfrac{24}{29}\) + \(\dfrac{34}{29}\)) + (\(\dfrac{32}{41}\) + \(\dfrac{50}{41}\))
= \(\dfrac{58}{29}\) + \(\dfrac{82}{41}\)
= 2 + 2
= 4
A=1+32+34+.....+32008
32A-A=1+32+34+....+32008+32010-[1+32+34+...+32008]
9A-A=32010
8A=32010
Mình làm vậy đúng hay sai.
A=1+32+33+34+......+32008
3A=3+33+34+35+......+32009
3A-A=(3+33+34+35.....+32009)-(1+32+33+34+...+32008)
A=(3+32009)-(1+32008)=(3+31+32008)-(1+32008)=3+3-1=5