2008*2009-1/2008^2+2017
Những câu hỏi liên quan
Tính nhanh
( 9/2010 + 10/2009 + 11/2008 - 1/2018 - 1/2017) ÷ ( 1/2019 + 1/2018 + 1/2017 - 1/2010 - 1/2009 - 1/2008)
1 : So sánh :
A = 2016/2017 + 2017/2018 và B = 2
A =2006/2007 + 2007/2008 +2008/2009 và B =3
19/18 và 2005/2004
A=2016/2017+2017/2018
Do 2016/2017<1,2017/2018<1=> A<2 Hay A<B
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so sánh 2008 với tổng 2009 số hạng sau\(s=\frac{2008+2007}{2009+2008}+\frac{^{2008^2+2007^2}}{2009^2+2008^2}+.....+\frac{2008^{2009}+2007^{2009}}{2009^{2009}+2008^{2009}}\)
So sánh \(A=\dfrac{2008^{2009}+2}{2008^{2009}-1};B=\dfrac{2008^{2009}}{2008^{2009}-3}\)
ta có: \(A=\dfrac{2008^{2009}+2}{2008^{2009}-1}=\dfrac{2008^{2009}-1+3}{2008^{2009}-1}=1+\dfrac{3}{2008^{2009}-1}\)
B=\(\dfrac{2008^{2009}}{2008^{2009}-3}=\dfrac{2008^{2009}-3+3}{2008^{2009}-3}=1+\dfrac{3}{2008^{2009}-3}\)
ta thấy: \(1+\dfrac{3}{2008^{2009}-1}\)<\(1+\dfrac{3}{2008^{2009}-3}\)
vậy A<B
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Tính nhanh:a)2/5+4/5x5/2
b)2008/2009-2009/2008+1/2009+2007/2008
a) \(\dfrac{2}{5}+\dfrac{4}{5}\times\dfrac{5}{2}\)
\(=\dfrac{2}{5}+\dfrac{4\times5}{5\times2}\)
\(=\dfrac{2}{5}+\dfrac{4}{2}\)
\(=\dfrac{2}{5}+2\)
\(=\dfrac{2}{5}+\dfrac{10}{5}\)
\(=\dfrac{12}{5}\)
b) \(\dfrac{2008}{2009}-\dfrac{2009}{2008}+\dfrac{1}{2009}+\dfrac{2007}{2008}\)
\(=\left(1-\dfrac{1}{2009}\right)-\left(1+\dfrac{1}{2008}\right)+\dfrac{1}{2009}+\left(1-\dfrac{1}{2008}\right)\)
\(=1-\dfrac{1}{2009}-1-\dfrac{1}{2008}+\dfrac{1}{2009}+1-\dfrac{1}{2008}\)
\(=\left(1-1+1\right)-\left(\dfrac{1}{2009}-\dfrac{1}{2009}\right)-\left(\dfrac{1}{2008}+\dfrac{1}{2008}\right)\)
\(=1-\dfrac{2}{2008}\)
\(=\dfrac{2008}{2008}-\dfrac{2}{2008}\)
\(=\dfrac{2006}{2008}\)
\(=\dfrac{1003}{1004}\)
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a: =2/5+4/2
=2/5+2
=12/5
b: \(=1-\dfrac{1}{2009}-1-\dfrac{1}{2008}+\dfrac{1}{2009}+1-\dfrac{1}{2008}\)
\(=1-\dfrac{2}{2008}=1-\dfrac{1}{1004}=\dfrac{1003}{1004}\)
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Tính nhanh :a)2/5+4/5 nhân 5/2
b)2008/2009-2009/2008+1/2009+2007/2008
so sánh 2 phân số : \(A=\frac{2008^{2009}+2}{2008^{2009}-1};B=\frac{2008^{2009}}{2008^{2009}-3}\)
SO SÁNH CÁC SỐ SAU
A=2008^2009+2/2008^2009-1 vàB=2008^2009/2008^2009-3
ta có A = 2008^2009+2 / 2008^2009-1 = 2008^2009-1+3 / 2008^2009-1 = 1 + 3/2008^2009-1
lại có B = 2008^2009 / 2008^2009-3 = 2008^2009-3+3 / 2008^2009-3 = 1 + 3/2008^2009-3
vì 3/2008^2009-1 < 3/2008^2009-3 => 1 + 3/2008^2009-1 < 1 + 3/2008^2009-3
Hay A<B
Vậy A<B
So sánh
bài 1 :A= 2006/2007-2007/2008+2008/2009-2009/2010
B= -1/2006*2007-1/2008*2009
bài 2: C= 2006/2007+2007/2008+2008/2009+2009/2006 với 4