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mình làm 2004.2004+2004+1002/(2004+1)(2004+1)-100... = 2004.2004+2004+1002/
2004.2004+2004+2004+1-1003 =
2004.2004+2004+1002/2004.2004+2004+1002
=1
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Sửa đề:
\(A=\dfrac{2004.2006+3006}{2005.2005-1003}\)
\(A=\dfrac{2004.2005+2004+3006}{2005.2005-1003}\)
\(A=\dfrac{2004.2005+2004+3006}{2004.2005+2005-1003}\)
\(A=\dfrac{2004.2005+5010}{2004.2005+1002}\)
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Ta có (2004.2004+3006)/(2005.2005-1003)=(2004.2004+2004+1002)/(2004+1).(2004+1)-1003=(2004.2004+2004+1002)/(2004.2004+2004+2004+1-1003)=(2004.2004+1002)/(2004.2004+1002)=1
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\(A=\dfrac{1995.1994-1}{1993.1995+1994}=\dfrac{1995\left(1993+1\right)-1}{1993.1995+1994}=\dfrac{1995.1993+1995-1}{1993.1995+1994}=\dfrac{1995.1993+1994}{1995.1993-1994}=1\)\(B=\dfrac{2004.2004+3006}{2005.2005-1003}=\dfrac{2004.2004+2004.1+1002}{2005.2005-1003}=\dfrac{2004.2005+1002}{2005.2005-1003}=\dfrac{2004.2005+1002}{2004.2005+2005-1003}=\dfrac{2004.2005+1002}{2004.2005+1002}=1\)\(C=\dfrac{2010.2011-1}{2009.2011+2010}=\dfrac{2009.2011+2011-1}{2009.2011+2010}=\dfrac{2019.2011+2010}{2009.20011+2010}=1\)\(D=\dfrac{2014.2015-1}{2013.2015+2013}=\dfrac{2013.2015+2014-1}{2013.2015+2013}=\dfrac{2013.2015+2013}{2013.2015+2013}=1\)
Câu 1 nhầm đề nha bạn mình sửa:
\(\dfrac{1995.1994-1}{1993.1995+1994}\)
\(=\dfrac{1995.\left(1993+1\right)-1}{1993.1995+1994}\)
\(=\dfrac{1995.1993+1995-1}{1993.1995+1994}\)
\(=\dfrac{1993.1995+1994}{1993.1995+1994}\)
\(=1\)
Câu 2: \(\dfrac{2004.2004+3006}{2005.2005-1003}\)
\(=\dfrac{2004.2004+2004+1002}{\left(2004+1\right).\left(2004+1\right)-1003}\)
\(=\dfrac{2004.2004+2004+1002}{2004.2004+2004+1-1003}\)
\(=\dfrac{2004.2004+2004+1002}{2004.2004+2004+1002}\)
\(=1\)
Câu 3:\(\dfrac{2010.2011-1}{2009.2011+2010}\)
\(=\dfrac{\left(2009+1\right).2011-1}{2009.2011+2010}\)
\(=\dfrac{2009.2011+2011-1}{2009.2011+2010}\)
\(=\dfrac{2009.2011+2010}{2009.2011+2010}\)
= 1
Câu 4:Nhầm để, sửa:
\(\dfrac{2014.2015-1}{2013.2015+2014}\)
\(=\dfrac{\left(2013+1\right).2015-1}{2013.2015+2014}\)
\(=\dfrac{2013.2015+2015-1}{2013.2015+2014}\)
\(=\dfrac{2013.2015+2014}{2013.2015+2014}\)
\(=1\)
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\(\frac{2004\times2004+3006}{2005\times2005-1003}=\frac{2004\times2004+3006}{\left(2004+1\right)\times\left(2004+1\right)-1003}\)
\(\frac{2004\times2004+3006}{2004\times\left(2004+1\right)+1\times\left(2004+1\right)-1003}=\frac{2004\times2004+3006}{2004\times2004+2\times2004+1-1003}\)
\(=\frac{2004\times2004+3006}{2004\times2004+3006}=1\)
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Làm 1 câu tặng 3 k
2 câu 6 k
3 câu 9 k
Nha
Please
a, 2 x 3 x 4 x 5 x 7 x 8 x 25 x 125 = 10 x 100 x 1000 x 3 x 7 = 21000000
Trả lời :
\(\frac{2004.2004+3006}{2005.2005-1003}\)\(=\)\(\frac{4016016+3006}{4020025-1003}\)
\(=\)\(\frac{4019022}{4019022}\)
\(=\)\(1\)