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Đặt A = 2003/1.2 + 2003/2.3 + 2003/3.4 + ... + 2003/2002.2003
A = 2003 . ( 1/1.2 + 1/2.3 + 1/3.4 + ... + 1/2002.2003 )
A = 2003 . ( 1 - 1/2003 )
A = 2003 . 2002/2003
A = 2002
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ee53eeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeee
S=<2003^1+2003^2+2003^3+2003^4+......+2003^10>
S+1=<2003.[1+2+3+...+10]>
S=2004.55
suy ra S:2004=55
vậy S chia hết cho 2004
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2002+2002.2+2002.3+2002.4+2003.5+2003.6
=2002.(1+2+3+4)+2003.(5+6)
=2002.10+2003.11
=2002.10+2003.10+2003
=10.(2002+2003)+2003
=10.4005+2003
=40050+2003
=42053
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A = \(\frac{2004-2003}{2004+2003}\)và B = \(\frac{2004^2-2003^2}{2004^2+2003^2}\)
Ta đặt : 2004 = x
2003 = y
Theo tính chất cơ bản của phân thức , ta có :
\(\frac{x-y}{x+y}=\frac{\left(x-y\right)\left(x+y\right)}{\left(x+y\right)\left(x+y\right)}=\frac{x^2-y^2}{x^2+y^2+2xy}\) ( 1 )
Vì x > 0 , y > 0 nên x2 + y2 + 2xy > x2 + y2
\(\Rightarrow\frac{x^2-y^2}{x^2+y^2+2xy}< \frac{x^2-y^2}{x^2+y^2}\) ( 2 )
Từ ( 1 ) và ( 2 )
\(\Rightarrow\frac{x-y}{x+y}< \frac{x^2-y^2}{x^2+y^2}\)
Vậy A < B
https://h.vn/hoi-dap/tim-kiem?q=so+s%C3%A1nh+2+ph%C3%A2n+s%E1%BB%91++A=+2004%5E2003++1+/+2004%5E2004++1++B=2004%5E2002+1/2004%5E2003++1&id=238505
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1 - 2 + 3 - 4 + ... + 99
= 1 + (-2 + 3) + (-4 + 5) + ... + (-98 + 99)
= 1 + 1 + 1 + ... + 1
= 1.50
= 50
(-2003) + (-21 + 75 + 2003)
= -2003 - 21 + 75 + 2003
= (-2003 + 2003) + (-21 + 75)
= 0 + 54
= 54
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Có:
- 2003A=20032004+2003/20032004+1 = 20032004+1+2002/20032004+1= 1+ 2002/20032004+1
- 2003A= 20032003+2003/20032003+1 .........= 1 + 2002/20032003+1
- Vì 1+ 2002/20032004+1<1+ 20022003+1nên 2003A<2003B
- Nên A<B
- !!!!!!!!!!!
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Đáp án của tớ là:
\(\frac{1}{1002}+\frac{1}{1003}+...+\frac{1}{2003}=\)\(\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2003}\right)-\)\(\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{1001}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2003}\right)-\)\(\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{2002}\right)-\)\(\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{2002}\right)=\)\(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2003}-\frac{1}{2}-\frac{1}{4}-\frac{1}{6}-...-\frac{1}{2002}\)\(-\frac{1}{2}-\frac{1}{4}-\frac{1}{6}-...-\frac{1}{2002}\)
Vậy:\(1+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2003}=\frac{1}{1002}+\frac{1}{1003}+...+\frac{1}{2003}\)
xin chòa hôm nay mình sẽ giúp bạn lam bài toán này
ta có
1/1002+1/1003+....+1/2003=(1+1/2+1/3+.....+1/2003)-(1+1/2+1/3+....+1/1001)
1/1002+1/1003+....+1/2003=(1+1/2+1/3+.....+1/2003)-(1/2+1/4+1/6+....+1/2002)-(1/2+1/4+1/6+......+1/2002)
1/1002+1/1003+.....+1/2003=1+1/2+1/3+....+1/2003-1/2+1/4+1/6+....+1/2002-1/2-1/4-1/6-....-1/2002
Vậy1/1002+1/1002+.....+1/2003=1-1/2+1/3-1/4+....-2/2002-1/2003
\(\left[\left(-2003\right)+\left(-250\right)\right]+275+2003=-2253+2278=25\)
\(\left(125.2^2-123.4\right):4=\left(125.4-123.4\right):4=4\left(125-123\right):4=2\)