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\(=\frac{2006.2004+2006-1}{2004.2006+2005}=\frac{2006.2004+2005}{2004.2006+2005}=1\)
\(a,\left(\frac{1}{2}\cdot\frac{1}{3}+\frac{1}{4}-\frac{1}{5}\right):\frac{1}{4}:\frac{1}{6}\)
\(=\left(\frac{1}{6}+\frac{1}{4}-\frac{1}{5}\right)\cdot\frac{1}{4}\cdot\frac{1}{6}\)
\(=\left(\frac{10}{60}+\frac{15}{60}-\frac{12}{60}\right)\cdot\frac{1}{24}\)
\(=\frac{13}{60}\cdot\frac{1}{24}\)
\(=\frac{13}{1440}\)
\(b,\frac{2006\cdot2005-1}{2004\cdot2006+2005}\)
\(\frac{2006\cdot2005-1}{2004\cdot2006+2005}\)
\(=\frac{2006\cdot\left(2004+1\right)-1}{2004 \cdot2006+2005}\)
\(=\frac{2006\cdot2004+2006\cdot1-1}{2004\cdot2006+2005}\)
\(=\frac{2006\cdot2004+2005}{2004\cdot2006 +2005}=1\)
Mình nghĩ phần b, ko có cách 2 đâu bạn .
\(\frac{2005.2007-1}{2004+2005.2006}\)
\(=\frac{2005.2006+2005-1}{2004+2005.2006}\)
\(=\frac{2005.2006+2004}{2004+2005.2006}\)
\(=1\)
\(=\frac{2015\left(2006+1\right)-1}{2004+2005.2006}=\frac{2005.2006+2005-1}{2004+2005.2006}=1\)
Ta có:
\(A=\frac{2003\times2004-1}{2003\times2004}=\frac{2003\times2004}{2003\times2004}-\frac{1}{2003\times2004}=1-\frac{1}{2003\times2004}\)
\(B=\frac{2004\times2005-1}{2004\times2005}=\frac{2004\times2005}{2004\times2005}-\frac{1}{2004\times2005}=1-\frac{1}{2004\times2005}\)
Vì \(\frac{1}{2003\times2004}>\frac{1}{2004\times2005}\Rightarrow A< B\)
Ta có:
2006x2005-1/2004x2006+2005 = 2006x(2004+1)-1 / 2004x2006+2005
= 2006x2004 + 2006 -1 / 2004x2006+2005
= 2006x2004+2005 / 2004x2006+2005
= 1
2006*2005-1/2004*2006+2005
=4022030 - 168/167 + 2005
=4022028,994 + 2005
=4024033 , 994
2006*2005-1/2004*2006+2005
=2006*2005-1/(2005-1)*2006+2005
=2006*2005-1/2005*2006-2006+2005
=2006*2005-1/2005*2006-1
=1
\(=\frac{2006x\left(2004+1\right)-1}{2004x2006+2005}=\frac{2006x2004+2005}{2004x2006+2005}=1\)
\(=\frac{2006\cdot2004+2006.1-1}{2004\cdot2006+2005}\)
\(=\frac{2006\cdot2004+2006-1}{2004\cdot2006+2005}\)
\(=\frac{2006\cdot2004+2005}{2004\cdot2006+2005}\)
\(\Rightarrow1\)
=2006×(2004+1)-1/2004×2006+2005
=2006×2004+2006×1-1/2004×2006+2005
=2006×2004+2005/2004×2006+2005
=1