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2.005x1004-8.02
= 2.005x1000+2.005x4-8.02
=2005+8.02-8.02
=2005
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\(\frac{2003\cdot1999-2003\cdot999}{2004\cdot999+1004}\)
\(=\frac{2003\cdot\left(1999-999\right)}{2004\cdot\left(999+1\right)}\)
\(=\frac{2003\cdot1000}{2004\cdot1000}\)
\(=\frac{2003}{2004}\)
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\(\dfrac{2003x1999-2003x999}{2004x999+1004}\)
\(=\dfrac{2003x\left(1999-999\right)}{2004x\left(1000-1\right)+1004}\)
\(=\dfrac{2003x1000}{2004x1000-2004+1004}\)
\(=\dfrac{2003x1000}{2004x1000-1000}\)
\(=\dfrac{2003x1000}{\left(2004-1\right)x1000}\)
\(=\dfrac{2003x1000}{2003x1000}=1\)
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\(∘Anya\)
\(=\dfrac{\text{1997 × 1996 − 995}}{\text{1995 × 1997 + 1002 }}\)
\(=\dfrac{\text{1997 × ( 1995 + 1 ) − 995}}{\text{1995 × 1997 + 1002}}\)
\(=\dfrac{\text{1997 × 1995 + 1997 − 995}}{\text{1995 × 1997 + 1002}}\)
\(=\dfrac{\text{1997 × 1995 + 1002}}{\text{1995 × 1997 + 1002 }}=1\)
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A=2003x2004-1/2003x2004
B=2004x2005-1/2004x2005
A= 1-2003x2004-1/2003x2004=1/2003x2004
B=1-2004x2005-1/2004x2005=1/2004x2005
Vì 1/2003x2004<1/2004x2005 => A>B.
K nhé
\(\frac{2002x2005-1002}{2003x2004-1004}\)
\(=\frac{2002x\left(2004+1\right)-1002}{\left(2002+1\right)x2004-1004}\)
\(=\frac{2002x2004+2002-1002}{2002x2004+2004-1004}\)
\(=\frac{2002x2004+1000}{2002x2004+1000}\)
\(=1\)
* Sai thì thông cảm nha!*
HT!~!