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Link dẫn:https://olm.vn/hoi-dap/question/230772.html
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- Ta có : `C=51/2 * 52/2 * 53/2* ... * 100/2`
`-> C=(51.52.53...100)/(2^50)`
`-> C=((1.2.3...50).(51.52.53...100))/((1.2.3...50).2^50)`
`-> C=(1.2.3...100)/((1.2).(2.2).(3.2)...(50.2))`
`-> C=(1.2.3...100)/(2.4.6...100)`
`-> C=1.3.5.7...99`
- Từ đó ta có :
`B-C=1.3.5.7...99-1.3.5.7...99=0`
- Vậy `B-C=0`
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\(C=\frac{\left(1.3.5.7...99\right).\left(2.4.6.8...98.100\right)}{2.4.6.8....98.100}=\frac{\left(1.2.3.4...99.100\right)}{\left(1.2\right).\left(2.2\right)\left(3.2\right)...\left(49.2\right).\left(50.2\right)}=\frac{1.2.3.4....99.100}{\left(1.2.3...49.50\right).\left(2.2.2...2\right)}\)
\(C=\frac{\left(1.2.3.4...50\right).\left(51.52.53...100\right)}{\left(1.2.3...50\right)\left(2.2.2....2\right)}=\frac{51}{2}.\frac{52}{2}...\frac{100}{2}=D\)
Vậy C = D
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Bài 2:b)Ta có:
D=(51*52*53*...*100):2^50.
=(51*53*55*...*99)*(52*54*56*...*100):2^50.
Khử 51*53*55*...*99 thì cần so sánh 1*3*5*...*41 với (52*54*56*...*100):2^50.
Lại có:
52*54*56*...*100:2^50=(52:2)*(54:2)*...*(100:2):(2^25) (vì 52;54;56;...;100 có 25 thừa số.
=26*27*28*...*50:2^25.
=(27*29*31*...*49)*(26*28*30*...*50):2^25
Khử với 1*3*5*...*49 thì cần so sánh 1*3*5*...*25 với (26*28*30*...*50):2^25.
Lại có:
26*28*30*...*50:2^25=(26:2)*(28:2)*(30:2)*...*(50:2):2^12(vì 26;28;30;...;50 có 13 thừa số).
=13*14*15*...*25:2^12.
=(13*15*17*19*21*23*25)*(14*16*18*20*22*24):2^12.
Khử với 1*3*5*...*25 thì cần so sánh 1*3*5*7*9*11 với (14*16*18*20*22*24):2^12.
Giờ số nhỏ rồi bấm máy tính so sánh là được.\
=>C=D.
Vậy C=D.
mấy câu kia dễ rồi tự l;àm nha mk nhắc câu khó thôi.
tk cho mk nha các bn.
-chúc ai tk mk học giỏi-
1/
a, x + (x+1) + (x+2) +...+ (x+100) = 2029099
(x+x+x+...+x) + (1+2+...+100) = 2029099
2011x + 2021055 = 2029099
2011x = 2029099 - 2021055
2011x = 8044
x = 8044 : 2011
x = 4
b, 2+4+6+....+2x = 210
=> 2(1+2+3+...+x) = 210
=> \(\frac{2x\left(x+1\right)}{2}=210\)
=> x(x+1) = 14.15
=> x = 14
2/
a, Vì B < 1
\(\Rightarrow B< \frac{2009^{2009}+1+2008}{2009^{2010}+1+2008}=\frac{2009^{2009}+2009}{2009^{2010}+2009}=\frac{2009\left(2009^{2008}+1\right)}{2009\left(2009^{2009}+1\right)}=\frac{2009^{2008}+1}{2009^{2009}+1}\)= A
Vậy A > B
b, Ta có:
\(D=\frac{51}{2}.\frac{52}{2}.\frac{53}{2}.....\frac{100}{2}=\frac{51.52.53....100}{2^{50}}\)
\(=\frac{\left(51.52.53....100\right)\left(1.2.3.4....50\right)}{2^{50}.\left(1.2.3.4....50\right)}\)
\(=\frac{1.2.3.4.5.6.....100}{\left(2.1\right)\left(2.2\right).\left(2.3\right).....\left(2.50\right)}\)
\(=\frac{1.2.3.4.5.6......100}{2.4.6........100}=\frac{\left(1.3.5....99\right)\left(2.4.6....100\right)}{2.4.6....100}\)
\(=1.3.5....99=C\)
Vậy C = D
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