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Đặt A= 2015^2013+1/2015^2014+7, B=2015^2014-2/2015^2015-2
2015A= 2015^2014+2015/2015^2014+7= 1 + (2008/2015^2014+7)
2015B= 2015^2015-4030/2015^2015-2= 1 - (4028/2015^2015-2)
Do 2015A>1>2015B nên A>B
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`A=(10^14-1)/(10^15-11)`
`=>10A=(10^15-10)/(10^15-11)`
`=>10A=(10^15-11+1)/(10^15-11)`
`=>10A=1+1/(10^15-1)`
`=>A>1/10`
`B=(10^14+1)/(10^15+9)`
`=>10B=(10^15+10)/(10^15+9)`
`=>10A=(10^15+9+1)/(10^15+9)`
`=>10A=1+1/(10^15+9)`
Vì `1/(10^15-1)>1/(10^15+9)`
`=>10B>10A`
`=>B>A`
Giải:
\(A=\dfrac{10^{14}-1}{10^{15}-11}\)
\(10A=\dfrac{10^{15}-10}{10^{15}-11}\)
\(10A=\dfrac{10^{15}-11+1}{10^{15}-11}\)
\(10A=1+\dfrac{1}{10^{15}-11}\)
Tương tự:
\(B=\dfrac{10^{14}+1}{10^{15}+9}\)
\(10B=\dfrac{10^{15}+10}{10^{15}+9}\)
\(10B=\dfrac{10^{15}+9+1}{10^{15}+9}\)
\(10B=1+\dfrac{1}{10^{15}+9}\)
Vì \(\dfrac{1}{10^{15}-11}>\dfrac{1}{10^{15}+9}\) nên \(10A>10B\)
\(\Rightarrow A>B\)
Chúc bạn học tốt!
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2014+2015+2016/2015+2016+2017<2014/2015+2015/2016+2016/2017
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\(\left(1-\frac{1}{1014}\right).\left(1-\frac{2}{1014}\right).\left(1-\frac{3}{1014}\right).\left(1-\frac{4}{1014}\right)...\left(1-\frac{1015}{1014}\right)\)
\(=\left(1-\frac{1}{1014}\right).\left(1-\frac{2}{1014}\right).\left(1-\frac{3}{1014}\right).\left(1-\frac{4}{1014}\right)...\left(1-\frac{1014}{1014}\right).\left(1-\frac{1015}{1014}\right)\)
\(=\left(1-\frac{1}{1014}\right).\left(1-\frac{2}{1014}\right).\left(1-\frac{3}{1014}\right).\left(1-\frac{4}{1014}\right)...\left(1-1\right).\left(1-\frac{1015}{1014}\right)\)
\(=\left(1-\frac{1}{1014}\right).\left(1-\frac{2}{1014}\right).\left(1-\frac{3}{1014}\right).\left(1-\frac{4}{1014}\right)...0.\left(1-\frac{1015}{1014}\right)\)
\(=0\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: \(\frac{2015}{-2014}=-\frac{2015}{2014}=-\left(1+\frac{1}{2014}\right)\)\(=-1-\frac{1}{2014}\)
\(\frac{-2016}{2015}=-\frac{2016}{2015}=-\left(1+\frac{1}{2015}\right)\)\(=-1-\frac{1}{2015}\)
Do \(\frac{1}{2014}\) > \(\frac{1}{2015}\) => \(\frac{2015}{-2014}\) < \(\frac{-2016}{2015}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có công thức :
\(\frac{a}{b}>\frac{a+c}{b+c}\)\(\left(\frac{a}{b}>1;a,b,c\inℕ^∗\right)\)
\(A=\frac{99^{2015}+1}{99^{2014}+1}>\frac{99^{2015}+1+98}{99^{2014}+1+98}=\frac{99^{2015}+99}{99^{2014}+99}=\frac{99\left(99^{2014}+1\right)}{99\left(99^{2013}+1\right)}=\frac{99^{2014}+1}{99^{2013}+1}=B\)
\(\Rightarrow\)\(A>B\)
Chúc bạn học tốt ~
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ta có A=1-\(\frac{1}{2014}\)
B=1-\(\frac{1}{2015}\)
Vì 2014 < 2015 => 1/2014 > 1/2015 => 1-1/2014 < 1-1/2015
Hay A<B
A = \(\frac{2013}{2014}=1-\frac{1}{2014}\) ; B = \(\frac{2014}{2015}=1-\frac{1}{2015}\). Vì \(\frac{1}{2014}>\frac{1}{2015}\)nên A<B
ta có: 1-(1014/1015)= 1/1015
1-(2014/2015)= 1/2015
vì 1/1015>1/2015 =>1014/1015<2014/2015
VẬY 1014/1015<2014/2015
có : 1-1014/1015=1/1015
1-2014/2015=1/2015
do 1/1015>1/2015
suy ra 1014/1015<2014/2015