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câu 1
125875/125871
câu 2
a) 13/17 < 15/19
b) 12/48 = 9/36
2006x125+1000 / 126x2005-888
= (2005+1)x125+1000 / (125+1)x2005-888
= 2005x125+125+1000 / 125x2005+2005-888
= 2005x125+1125 / 125x2005+1117
= 1125/1117
\(\frac{2006x125+1000}{126x2006-1006}=\frac{2006x125+1000}{125x2006-2006-1006}=\frac{2006x125+1000}{125x2006+1000}=1\)
=\(\frac{2006x125+1000}{125x2006+2006-1006}\)
=\(\frac{2006x125+1000}{125x2006+1000}\)
=1
\(\frac{2006\cdot125+1000}{126\cdot2006-1006}=\frac{2006\cdot\left(126-1\right)+1000}{126\cdot2006-1006}\)
\(=\frac{2006\cdot126-2006+1000}{126\cdot2006-1006}\)
\(=\frac{2006\cdot126-1006}{126\cdot2006-1006}\)
\(=1\)
\(\frac{2006x125+1000}{126x2006-1006}\)
\(=\frac{2006x125+1000}{\left(125+1\right)x2006-1006}\)
\(=\frac{2006x125+1000}{125x2006+1x2006-1006}\)
\(=\frac{2006x125+1000}{125x2006+2006-1006}\)
\(=\frac{2006x125+1000}{125x2006+1000}\)
\(=1\)
\(\frac{\left(2005+1\right).125+1000}{\left(125+1\right).2005-888}\)
= \(\frac{2005.125+125+1000}{125.2005+2005-888}\)
= \(\frac{2005.125+1125}{125.2005+1117}\)
= \(\frac{250625+1125}{250625+1117}\)
= \(\frac{125875}{125871}\)
A = 2006 x 125 + 1000 / 126 x 2005 - 888
A = (2005 + 1) x 125 + 1000 / (125 + 1) x 2005 - 888
A = 2005 x 125 + 125 + 1000 / 125 x 2005 + 2005 - 888
A = 2005 x 125 + 1125 / 125 x 2005 + 1117
A = 1125 / 1117