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\(\dfrac{2007\times2006-1}{2005\times2007+2006}\)

\(=\dfrac{2007\times2006-1}{\left(2006-1\right)\times2007+2006}\)

\(=\dfrac{2007\times2006-1}{2006\times2007-2007+2006}\)

\(=\dfrac{2007\times2006-1}{2006\times2007-\left(2007-2006\right)}\)

\(=\dfrac{2007\times2006-1}{2007\times2006-1}\)

\(=1\)

\(\dfrac{2007x2006-1}{2005x2007+2006}\)

\(=\dfrac{2007x\left(2005+1\right)-1}{2005x2007+2006}\)

\(=\dfrac{2007x2005+2007-1}{2005x2007+2006}\)

\(=\dfrac{2007x2005+2006}{2005x2007+2006}=1\)

1

2003 / 2001 = 1 + 2/2001

1999/1997 = 1 + 2/1997 

vì 2/ 2001 < 2/1997

nên 1 + 2/2001 < 1 + 2/1997

hay 2003 < 1999/1997

b, = 5/9 x 1/4 + 4/9 x 1/4 

= 1/4 x ( 5/9 + 4/9 )

= 1/4 x 1 

= 1/4

* Ý a mk k nhớ cách làm ^^, xl * 

\(b,\dfrac{5}{9}\times\dfrac{1}{4}+\dfrac{4}{9}\times\dfrac{3}{12}\)

\(=\dfrac{5}{9}\times\dfrac{1}{4}+\dfrac{4}{9}\times\dfrac{1}{4}\)

\(=\dfrac{1}{4}\times\left(\dfrac{5}{9}+\dfrac{5}{9}\right)\)

\(=\dfrac{1}{4}\times\dfrac{9}{9}=\dfrac{1}{4}\times1=\dfrac{1}{4}\)

Giúp mình, mình đang cần gấp!!!

có phải 0 ko

C=(2005x2006+2005-1)/(2004+2005x2006)=

(2005x2006+2004)/(2004+2005x2006)=1

rồi cảm ơn tôi đã làm trước khi cậu ra đáp án

Bài 1:

+) \(\dfrac{7}{8}\times y=\dfrac{3}{2}+\dfrac{6}{4}=3\)

\(y=3:\dfrac{7}{8}=\dfrac{24}{7}\)

+) \(\dfrac{1}{y}\times\left(\dfrac{2}{5}+\dfrac{1}{5}\right)=\dfrac{10}{3}\)

\(\dfrac{1}{y}=\dfrac{10}{3}:\dfrac{3}{5}=\dfrac{50}{9}\)

\(y=\dfrac{9}{50}\)

Bài 2:

+) \(=\dfrac{2}{5}\times\left(\dfrac{4}{7}+\dfrac{3}{7}\right)\)

\(=\dfrac{2}{5}\times\dfrac{7}{7}=\dfrac{2}{5}\)

+) \(\dfrac{2}{9}:\dfrac{2}{3}:\dfrac{3}{9}\)

\(\dfrac{2}{9}\times\dfrac{3}{2}\times\dfrac{9}{3}=1\)

viết rứa ai mà biết

\(=\dfrac{-1}{2}\cdot\dfrac{-2}{3}\cdot...\cdot\dfrac{-1998}{1999}\)

\(=\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot...\cdot\dfrac{1998}{1999}=\dfrac{1}{1999}\)

\(\left(\dfrac{1}{2}-1\right)\left(\dfrac{1}{3}-1\right)\left(\dfrac{1}{4}-1\right)...\left(\dfrac{1}{1999}-1\right)\)

\(=\left(\dfrac{1}{2}-\dfrac{2}{2}\right)\cdot\left(\dfrac{1}{3}-\dfrac{3}{3}\right)\cdot\left(\dfrac{1}{4}-\dfrac{4}{4}\right)...\left(\dfrac{1}{1999}-\dfrac{1999}{1999}\right)\)

\(=\dfrac{1-2}{2}\cdot\dfrac{1-3}{3}\cdot\dfrac{1-4}{4}\cdot...\cdot\dfrac{1-1999}{1999}\)

\(=\dfrac{-1}{2}\cdot\dfrac{-2}{3}\cdot\dfrac{-3}{4}\cdot...\cdot\dfrac{-1998}{1999}\)

\(=\dfrac{-1\cdot-2\cdot-3\cdot...\cdot-1998}{2\cdot3\cdot4\cdot...\cdot1999}\)

\(=\dfrac{1}{1999}\)

     C =           \(\dfrac{1}{2}\) + \(\dfrac{1}{4}\) + \(\dfrac{1}{8}\) + \(\dfrac{1}{16}\) + \(\dfrac{1}{32}\) + \(\dfrac{1}{64}\) + \(\dfrac{1}{128}\)

  2\(\times\)C =   1 +  \(\dfrac{1}{2}\)  + \(\dfrac{1}{4}\) + \(\dfrac{1}{8}\) + \(\dfrac{1}{16}\) + \(\dfrac{1}{32}\) + \(\dfrac{1}{64}\) 

2 \(\times\) C - C =   1 -  \(\dfrac{1}{128}\)

       C       = \(\dfrac{127}{128}\)

uh là nhân

a) \(=\dfrac{254x\left(400-1\right)-145}{254+\left(400-1\right)x253}=\dfrac{254x400-254-145}{254+253x400-253}\)

\(=\dfrac{101600-399}{101200+1}=\dfrac{101211}{101201}=\dfrac{101201+10}{101201}=1+\dfrac{10}{101201}\)

b) \(=\dfrac{5392+\left(600+1\right)x5391}{5392x\left(600+1\right)-69}=\dfrac{5392+600x5391+5391}{5392x600+5392-69}\)

\(=\dfrac{10783+3234600}{3235200+5323}=\dfrac{\text{3245383}}{\text{3240523}}=\dfrac{3240523+60}{3240523}=1+\dfrac{60}{3240523}\)

c) \(=\dfrac{1}{2}x\left(\dfrac{1}{2}-\dfrac{1}{4}\right)+\dfrac{1}{2}x\left(\dfrac{1}{4}-\dfrac{1}{8}\right)+\dfrac{1}{32}\)

\(=\dfrac{1}{2}x\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{8}\right)+\dfrac{1}{32}\)

\(=\dfrac{1}{2}x\left(\dfrac{1}{2}-\dfrac{1}{8}\right)+\dfrac{1}{32}=\dfrac{3}{16}+\dfrac{1}{32}=\dfrac{7}{32}\)