Bài 4: Tính nhanh:
\(\dfrac{2007x2006-8}{2005x2007+1999}\)
K
Khách
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H9
HT.Phong (9A5)
CTVHS
1 tháng 9 2023
\(\dfrac{2007\times2006-1}{2005\times2007+2006}\)
\(=\dfrac{2007\times2006-1}{\left(2006-1\right)\times2007+2006}\)
\(=\dfrac{2007\times2006-1}{2006\times2007-2007+2006}\)
\(=\dfrac{2007\times2006-1}{2006\times2007-\left(2007-2006\right)}\)
\(=\dfrac{2007\times2006-1}{2007\times2006-1}\)
\(=1\)
ND
Nguyễn Đức Trí
VIP
1 tháng 9 2023
\(\dfrac{2007x2006-1}{2005x2007+2006}\)
\(=\dfrac{2007x\left(2005+1\right)-1}{2005x2007+2006}\)
\(=\dfrac{2007x2005+2007-1}{2005x2007+2006}\)
\(=\dfrac{2007x2005+2006}{2005x2007+2006}=1\)
8 tháng 5 2022
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
2
8 tháng 5 2022
* Ý a mk k nhớ cách làm ^^, xl *
\(b,\dfrac{5}{9}\times\dfrac{1}{4}+\dfrac{4}{9}\times\dfrac{3}{12}\)
\(=\dfrac{5}{9}\times\dfrac{1}{4}+\dfrac{4}{9}\times\dfrac{1}{4}\)
\(=\dfrac{1}{4}\times\left(\dfrac{5}{9}+\dfrac{5}{9}\right)\)
\(=\dfrac{1}{4}\times\dfrac{9}{9}=\dfrac{1}{4}\times1=\dfrac{1}{4}\)
PH
15 tháng 3 2015
C=(2005x2006+2005-1)/(2004+2005x2006)=
(2005x2006+2004)/(2004+2005x2006)=1
MH
9 tháng 5 2022
Bài 1:
+) \(\dfrac{7}{8}\times y=\dfrac{3}{2}+\dfrac{6}{4}=3\)
\(y=3:\dfrac{7}{8}=\dfrac{24}{7}\)
+) \(\dfrac{1}{y}\times\left(\dfrac{2}{5}+\dfrac{1}{5}\right)=\dfrac{10}{3}\)
\(\dfrac{1}{y}=\dfrac{10}{3}:\dfrac{3}{5}=\dfrac{50}{9}\)
\(y=\dfrac{9}{50}\)
29 tháng 8 2023
\(=\dfrac{-1}{2}\cdot\dfrac{-2}{3}\cdot...\cdot\dfrac{-1998}{1999}\)
\(=\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot...\cdot\dfrac{1998}{1999}=\dfrac{1}{1999}\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
18 tháng 8 2023
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 = \(\dfrac{127}{128}\)
ND
Nguyễn Đức Trí
VIP
14 tháng 7 2023
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) \(=\dfrac{1}{2}x\left(\dfrac{1}{2}-\dfrac{1}{4}\right)+\dfrac{1}{2}x\left(\dfrac{1}{4}-\dfrac{1}{8}\right)+\dfrac{1}{32}\)
\(=\dfrac{1}{2}x\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{8}\right)+\dfrac{1}{32}\)
\(=\dfrac{1}{2}x\left(\dfrac{1}{2}-\dfrac{1}{8}\right)+\dfrac{1}{32}=\dfrac{3}{16}+\dfrac{1}{32}=\dfrac{7}{32}\)
\(\dfrac{2006\times2007-8}{2005\times2007+1999}\)
= \(\dfrac{\left(2005+1\right)\times2007-8}{2005\times2007+1999}\)
= \(\dfrac{2005\times2007+2007-8}{2005\times2007+1999}\)
= \(\dfrac{2005\times2007+1999}{2005\times2007+1999}\)
= 1