Thực hiện phép tính:
(\(\dfrac{2003}{2004}+\dfrac{2004}{2003}\)): \(\dfrac{8028045}{8028024}\)
*Giúp mình với, mình đang cần gấp lắm ạ
Những câu hỏi liên quan
Bài 1: Thực hiện phép tính:
a) ( 2- dfrac{3}{2}). ( 2- dfrac{4}{3}). (2- dfrac{5}{4}). ( 2- dfrac{6}{5})
b) dfrac{1}{2002}+dfrac{2003.2001}{2002}-2003
c)( dfrac{2003}{2004}+dfrac{2004}{2003}):dfrac{8028045}{8028024}
d) 4+ dfrac{1}{1+dfrac{1}{1+dfrac{2}{1+dfrac{3}{4}}}}
Đọc tiếp
Bài 1: Thực hiện phép tính:
a) ( 2- \(\dfrac{3}{2}\)). ( 2- \(\dfrac{4}{3}\)). (2- \(\dfrac{5}{4}\)). ( 2- \(\dfrac{6}{5}\))
b) \(\dfrac{1}{2002}+\dfrac{2003.2001}{2002}-2003\)
c)( \(\dfrac{2003}{2004}+\dfrac{2004}{2003}\)):\(\dfrac{8028045}{8028024}\)
d) 4+ \(\dfrac{1}{1+\dfrac{1}{1+\dfrac{2}{1+\dfrac{3}{4}}}}\)
tính :
a, (2-dfrac{3}{2} ) . (2-dfrac{4}{3} ) . (2-dfrac{5}{4} ) . (2-dfrac{6}{5} )
b,dfrac{1}{2002} + dfrac{2003.2001}{2002} - 2003
c, (dfrac{2003}{2004} + dfrac{2004}{2003} ) : dfrac{8028025}{8028024}
giúp mình với mình đang cần gấp
Đọc tiếp
tính :
a, (2-\(\dfrac{3}{2}\) ) . (2-\(\dfrac{4}{3}\) ) . (2-\(\dfrac{5}{4}\) ) . (2-\(\dfrac{6}{5}\) )
b,\(\dfrac{1}{2002}\) + \(\dfrac{2003.2001}{2002}\) - 2003
c, (\(\dfrac{2003}{2004}\) + \(\dfrac{2004}{2003}\) ) : \(\dfrac{8028025}{8028024}\)
giúp mình với mình đang cần gấp
a, \(\left(2-\dfrac{3}{2}\right)\left(2-\dfrac{4}{3}\right)\left(2-\dfrac{5}{4}\right)\left(2-\dfrac{6}{5}\right)\)
\(=\left(\dfrac{4}{2}-\dfrac{3}{2}\right)\left(\dfrac{6}{3}-\dfrac{4}{3}\right)\left(\dfrac{8}{4}-\dfrac{5}{4}\right)\left(\dfrac{10}{5}-\dfrac{6}{5}\right)\)
\(=\dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}.\dfrac{4}{5}\)
\(=\dfrac{1}{5}\)
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b. \(\dfrac{1}{2002}+\dfrac{2003.2001}{2002}-2003\)\(=\dfrac{1}{2002}+\dfrac{2003.2001}{2002}-\dfrac{2003.2002}{2002}\) = \(\dfrac{1+2003.2001-2003.2002}{2002}\) = \(\dfrac{1+\left(2003\left(2001-2002\right)\right)}{2002}\) = \(\dfrac{1+2003.\left(-1\right)}{2002}\) = \(\dfrac{1+\left(-2003\right)}{2002}\) = \(\dfrac{-2002}{2002}=-1\)
Chúc nguyễn hồng nhung học tốt
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giúp mình câu này với:
tính N=2003(2004^9+2004^8+...+2004^2+2005)+1
mình cần gấp lắm
CMR \(8^{2003}+5^{2003}+17^{2004}-4^{2004}⋮13\)
làm ơn hãy giúp mình làm bài này mình đang cần gấp lắm
so sánh M,N
M=\(\dfrac{2003}{2004}+\dfrac{2004}{2005}\)
N=\(\dfrac{2003+2004}{2004+2005}\)
Ta có:
N=\(\dfrac{2003+2004}{2004+2005}\)=\(\dfrac{2003}{2004+2005}\)+\(\dfrac{2004}{2004+2005}\)
Ta thấy:
\(\dfrac{2003}{2004+2005}\)<\(\dfrac{2003}{2004}\)(1)
\(\dfrac{2004}{2004+2005}\)<\(\dfrac{2004}{2005}\)(2)
Từ (1) và (2) --> M=\(\dfrac{2003}{2004}\)+\(\dfrac{2004}{2005}\)>\(\dfrac{2003}{2004+2005}\)+\(\dfrac{2004}{2004+2005}\)=N
Vậy M>N
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I, Tìm x: a, \(\dfrac{x-2004}{2003}+\dfrac{x-2003}{2005}+\dfrac{x-2005}{2004}=3+\dfrac{2005}{2004}+\dfrac{2004}{2005}\)
Tính nhanh giúp mình với
127^2+146×126+73^2
Mình cần gấp lắm ạ
2006^2-2005^2+2004^2-2003^2+...+2^2-1^2
Giúp mình 1 câu thôi cũng được ạ
127^2 + 146 x 126 + 73^2
= 127^2 + 2 x 73 x 126 + 73 x 73
= 127^2 + 73 x ( 2 x126 + 73 )
=......
rồi sau đo tinh binh thuong mk chi co the giup vay thoi
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Thực hiện các phép tính sau:
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a)\(\left(2-\frac{3}{2}\right).\left(2-\frac{4}{3}\right).\left(2-\frac{5}{4}\right).\left(2-\frac{6}{4}\right)\)
b) \(\left(\frac{2003}{2004}+\frac{2004}{2003}\right):\frac{8028025}{8028024}\)
a) \(\left(2-\frac{3}{2}\right)\left(2-\frac{4}{3}\right)\left(2-\frac{5}{4}\right)\left(2-\frac{6}{4}\right)\)
\(=\frac{1}{3}\left(-\frac{4}{3}+2\right)\left(-\frac{5}{4}+2\right)\left(-\frac{6}{4}+2\right)\)
\(=\frac{1}{2}.\frac{2}{3}\left(-\frac{5}{4}+2\right)\left(-\frac{6}{4}+2\right)\)
\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}\left(-\frac{6}{4}+2\right)\)
\(=\frac{1.2.3\left(2-\frac{3}{2}\right)}{2.3.4}\)
\(=\frac{1.3\left(2-\frac{3}{2}\right)}{3.4}\)
\(=\frac{1.\left(2-\frac{3}{2}\right)}{4}\)
\(=\frac{2-\frac{3}{4}}{4}\)
\(=\frac{1}{2.4}\)
\(=\frac{1}{8}\)
b) \(\left(\frac{2003}{2004}+\frac{2004}{2003}\right):\frac{8028025}{8028024}\)
\(=\frac{8028024\left(\frac{2003}{2004}+\frac{2004}{2003}\right)}{8028025}\)
\(=\frac{8028024.\frac{8028025}{4014012}}{8028025}\)
\(=\frac{16056050}{8028025}\)
= 2
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so sánh
\(\dfrac{2003}{2004}+\dfrac{2004}{2005}\) và \(\dfrac{2003+2004}{2004+2005}\)
Cho A=\(\dfrac{2003}{2004}\)+\(\dfrac{2004}{2005}\); B=\(\dfrac{2003+2004}{2004+2005}\)
Ta có: B=\(\dfrac{2003}{2004+2005}\)+\(\dfrac{2004}{2004+2005}\)
Vì: \(\dfrac{2003}{2004+2005}< \dfrac{2003}{2004}\)
\(\dfrac{2004}{2004+2005}< \dfrac{2004}{2005}\)
=>\(\dfrac{2003}{2004+2005}+\dfrac{2004}{2004+2004}< \dfrac{2003}{2004}+\dfrac{2004}{2005}\)
=>\(\dfrac{2003+2004}{2004+2005}< \dfrac{2003}{2004}+\dfrac{2004}{2005}\)
=>B<A
Vậy B<A
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