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\(=\frac{2006.2004+2006-1}{2004.2006+2005}=\frac{2006.2004+2005}{2004.2006+2005}=1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
=2006×(2004+1)-1/2004×2006+2005
=2006×2004+2006×1-1/2004×2006+2005
=2006×2004+2005/2004×2006+2005
=1
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A=4024035-1/2004+2005*2006
=4024035+2005*2006
=4024035+4011
=4028046
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(1/2003+1/2004+1/2005) / (2/2003+2/2004+2/2005)
= (1/2003+1/2004+1/2005) / 2(1/2003+1/2004+1/2005)
= 1/2
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\(a,\left(\frac{1}{2}\cdot\frac{1}{3}+\frac{1}{4}-\frac{1}{5}\right):\frac{1}{4}:\frac{1}{6}\)
\(=\left(\frac{1}{6}+\frac{1}{4}-\frac{1}{5}\right)\cdot\frac{1}{4}\cdot\frac{1}{6}\)
\(=\left(\frac{10}{60}+\frac{15}{60}-\frac{12}{60}\right)\cdot\frac{1}{24}\)
\(=\frac{13}{60}\cdot\frac{1}{24}\)
\(=\frac{13}{1440}\)
\(b,\frac{2006\cdot2005-1}{2004\cdot2006+2005}\)
\(\frac{2006\cdot2005-1}{2004\cdot2006+2005}\)
\(=\frac{2006\cdot\left(2004+1\right)-1}{2004 \cdot2006+2005}\)
\(=\frac{2006\cdot2004+2006\cdot1-1}{2004\cdot2006+2005}\)
\(=\frac{2006\cdot2004+2005}{2004\cdot2006 +2005}=1\)
Mình nghĩ phần b, ko có cách 2 đâu bạn .
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\(\frac{2005\cdot2004-1}{2003\cdot2005+2004}\)
\(=\frac{2005\cdot\left(2003+1\right)-1}{2003\cdot2005+2004}\)
\(=\frac{2005\cdot2003+2005-1}{2003\cdot2005+2004}\)
\(=\frac{2005\cdot2003+2004}{2003\cdot2005+2004}\)
\(=1\)
2005 x 2004 - 1 / 2003 × 2005 + 2004
= 2005 × (2003 + 1) - 1 / 2003 × 2005 + 2004
= 2005 × 2003 + (2005 - 1) / 2003 × 2005 + 2004
= 2005 × 2003 + 2004 / 2003 × 2005 + 2004
= 1
\(\frac{2005.2007-1}{2004+2005.2006}\)
\(=\frac{2005.2006+2005-1}{2004+2005.2006}\)
\(=\frac{2005.2006+2004}{2004+2005.2006}\)
\(=1\)
\(=\frac{2015\left(2006+1\right)-1}{2004+2005.2006}=\frac{2005.2006+2005-1}{2004+2005.2006}=1\)