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ta có
\(B=1+\left(1-\frac{1}{2}\right)+..+\left(1-\frac{1}{100}\right)\)
\(=1+\frac{1}{2}+\frac{2}{3}+..+\frac{99}{100}=A\)
Vậy A=B
\(1+2-3-4+5+6-7-8+...-300+301\)
\(=1+\left(2-3-4+5\right)+\left(6-7-8+9\right)+...+\left(298-299-300+301\right)\)
\(=1+0+0+...+0\)
\(=1\)
A = 1 + 2 - 3 - 4 + 5 + 6 - 7 - 8 + 9 + 10 -...- 299 - 300 + 301 + 302
A =1+(2 -3 - 4 + 5) + (6 - 7 - 8 + 9) +....+ (298 - 299 - 300 + 301)+ 302
A = 1 + 0 +....+ 0 + 302
A = 303
mik mới so sánh được với 98 thôi. Mik ra kq là A > 99 - 99/100 -> A > 98 chứ chưa so sánh đc với 99.
`A=3/4+8/9+.............+9999/10000`
`=1-1/4+1-1/9+,,,,,,,,,,+1-1/10000`
`=99-(1/4+1/9+.........+1/10000)<99-0=99`
`=>A<99`
A = 1 . 3 + 3 . 5 + 5 . 7 + ... + 49 . 51
= 1 . 51
= 51
B = 2 . 4 + 4 . 6 + 6 . 8 + ... + 98 . 100
= 2 . 100
= 200
C = 1 . 4 + 4 . 7 + 7 . 10 + ... + 301 . 304
= 1 . 304
= 304
D = 1 + 1 . 1! + 2 . 2! + 3 . 3! + ... + 100 . 100!
= 1 . 100
= 100
E = 22 + 42 + ... + ( 2n )2
= 22 . ( 2n )2
= 2n4
`3/(-10) ; 1/(-2) ; 4/(-5)=> -3/10 ; -1/2 ; -4/5`
ta có : `-1/2=(-1xx5)/(2xx5)=-5/10 ; -4/5=(-4xx2)/(5xx2)=-8/10`
vậy `3/(-10) < 1/(-2) < 4/(-5)`
`--------------------`
`2/(-10) ; 7/(-5) ; -1/2=>-2/10 ;-7/5;-1/2`
ta có : `-7/5=(-7xx2)/(5xx2)=-14/10; -1/2=(-1xx5)/(2xx5)=-5/10`
vậy `2/(-10) < -1/2 < 7/(-5)`
`---------------------`
`7/(-4) ; -2/5 ; -3/10=> -7/4;-2/5;-3/10`
ta có : `-7/4=(-7xx5)/(4xx5)=-35/20 ; -2/5=(-2xx4)/(5xx4)=-8/20;-3/10=(-3xx2)/(10xx2)=-6/20`
vậy 7/(-4) > -2/5 > -3/10`
a) 17.13+17.42-17.35
=17.(13+42-35)
=17.20=340
b) [25.(18-42)-10]:4+6
=(25.2-10):4+6
=40:4+6=16
c) 36:32+23.22-32.3
=34+25-33
=81+32-27=86
d) B=3.42-22.3
=3.(16-4)
=3.12=36
e)20220+3.[52.10-(23-13)2]
=1+3.(250-100)
=1+450=451
g) 27.77+24.27-27
=27.(77+24-1)
=27.100=2700
h) 5.23+79:77-12020
=40+72-1
=89-1=88
i) 120:{54[50:2+(32-2.4)]}
=120:[54(25+1)]
=120:1404=10/117
\(E=\dfrac{4}{3}+\dfrac{7}{3^2}+\dfrac{10}{3^3}+....+\dfrac{298}{3^{99}}+\dfrac{301}{3^{100}}\)
\(3E=4+\dfrac{7}{3}+\dfrac{10}{3^2}+....+\dfrac{298}{3^{98}}+\dfrac{301}{3^{99}}\)
\(3E-E=4+\dfrac{3}{3}+\dfrac{3}{3^2}+...+\dfrac{3}{3^{98}}+\dfrac{3}{3^{99}}-\dfrac{301}{3^{100}}\)
\(2E=1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{97}}+\dfrac{1}{3^{98}}+4-\dfrac{301}{3^{100}}\)
Đặt
\(A=1+\dfrac{1}{3}+...+\dfrac{1}{3^{97}}+\dfrac{1}{3^{98}}\)
\(3A=3+1+\dfrac{1}{3}+....+\dfrac{1}{3^{96}}+\dfrac{1}{3^{97}}\)
\(3A-A=3-\dfrac{1}{3^{98}}\)
\(2A=3-\dfrac{1}{3^{98}}\)
\(A=\dfrac{3}{2}-\dfrac{1}{3^{98}\times2}\)
\(\Rightarrow2E=\dfrac{3}{2}-\dfrac{1}{3^{98}\times2}+4-\dfrac{301}{3^{100}}\)
\(2E=\dfrac{11}{2}-\dfrac{1}{3^{98}\times2}-\dfrac{301}{3^{100}}\)
\(\Rightarrow2E< \dfrac{11}{2}\Rightarrow E< \dfrac{11}{4}=2,75\)