so sánh cách nhanh nhất :
3/2006 + 7/2017 và 7/2006 + 3/2017
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2017*4+2017*62+2017*35-2017
=2017*(4+62+35-1)
=2017*100
=201700
2006*42+2006*39+2006*21-2006*2
=2006*(42+39+21-2)
=2006*100
=200600
k mik đi
Ta có : \(\frac{2007}{2006}>1\)\(;\)\(\frac{2016}{2017}< 1\)
\(=>\frac{2007}{2006}>\frac{2016}{2017}\)
2017>2016 nên 2017:2016 >1
Vì 2016<2017 nên 2016:2017<1
Suy ra 2017:2016<2016:2017
A=2016/2017+2017/2018
Do 2016/2017<1,2017/2018<1=> A<2 Hay A<B
a)có tử và mẫu lớn hơn thì lớn hơn!
b)có tử,mẫu lớn hơn thì bé hơn!
c)tương tự câu a
đúng 100%
A = \(\frac{-7}{10^{2005}}\)+ \(\frac{-8}{10^{2006}}\)+ \(\frac{-7}{10^{2006}}\)
B = \(\frac{-7}{10^{2005}}\)+ \(\frac{-8}{10^{2005}}\)+ \(\frac{-7}{10^{2006}}\)
Vì \(\frac{-7}{10^{2005}}\)= \(\frac{-7}{10^{2005}}\); \(\frac{-7}{10^{2006}}\)= \(\frac{-7}{10^{2006}}\); \(\frac{-8}{10^{2006}}\)> \(\frac{-8}{10^{2005}}\) ( vì tử chung là số âm nên mẫu lớn hơn thì phân số đó lớn hơn)
=> \(\frac{-7}{10^{2005}}\)+ \(\frac{-7}{10^{2006}}\)+ \(\frac{-8}{10^{2006}}\)> \(\frac{-7}{10^{2005}}\)+ \(\frac{-7}{10^{2006}}\)+ \(\frac{-8}{10^{2005}}\)
=> A > B
\(T=\frac{2}{2^1}+\frac{3}{2^2}+\frac{4}{2^3}+...+\frac{2017}{2^{2016}}\) => \(\frac{T}{2}=\frac{2}{2^2}+\frac{3}{2^3}+\frac{4}{2^4}+...+\frac{2017}{2^{2017}}\)
=> \(T-\frac{T}{2}=\left(\frac{2}{2^1}+\frac{3}{2^2}+\frac{4}{2^3}+...+\frac{2017}{2^{2016}}\right)-\left(\frac{2}{2^2}+\frac{3}{2^3}+\frac{4}{2^4}+...+\frac{2017}{2^{2017}}\right)\)
<=> \(\frac{T}{2}=\frac{2}{2^1}+\left(\frac{3}{2^2}-\frac{2}{2^2}\right)+\left(\frac{4}{2^3}-\frac{3}{2^3}\right)+...+\left(\frac{2017}{2^{2016}}-\frac{2016}{2^{2016}}\right)-\frac{2017}{2^{2017}}\)
<=> \(\frac{T}{2}=1+\left(\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2016}}\right)-\frac{2017}{2^{2017}}\)
Đặt: \(M=\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2016}}=>2M=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2015}}\)
=> \(2M-M=\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2015}}\right)-\left(\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2016}}\right)\)
=> \(M=\frac{1}{2}-\frac{1}{2^{2016}}< \frac{1}{2}\)
=> \(\frac{T}{2}< 1+\frac{1}{2}-\frac{2017}{2^{2017}}< 1+\frac{1}{2}=\frac{3}{2}\)
=> T < 3
3/2006 + 7/2017 = 3/2006 + (3/2017 + 4/2017) = 3/2006 + 3/2017 + 4/2017
7/2006 + 3/2017 = (3/2006 + 4/2006) + 3/2017 = 3/2006 + 4/2006 + 3/2017
Vì 4/2006 > 4/2017 => 3/2006 + 3/2017 + 4/2017 < 3/2006 + 4/2006 + 3/2017
=> 3/2006 + 7/2017 < 7/2006 + 3/2017