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![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có : \(A=\left(\left(-2015\right)^{2016}.-2016^{2017}+\left(-2016\right)^{2017}.-2015^{2016}\right).\left(-2017\right)^{2018}\)
\(=\left(2015^{2016}.-2016^{2017}-2016^{2017}.-2015^{2016}\right).2017^{2018}\)
\(=\left(2015^{2016}-2015^{2016}\right).2017^{2018}.\left(-2016^{2017}\right)\)
\(=0.2017^{2018}.\left(-2016^{2017}\right)=0\)
Giải:
\(A=\left[\left(-2015\right)^{2016}.\left(-2016^{2017}\right)+\left(-2016\right)^{2017}.\left(-2015^{2016}\right)\right].\left(-2017\right)^{2018}\)
\(A=\left[2015^{2016}.\left(-2016\right)^{2017}+\left(-2016\right)^{2017}.\left(-2015^{2016}\right)\right].\left(-2017\right)^{2018}\)
\(A=\left[2015^{2016}+\left(-2015^{2016}\right)\right].\left(-2016\right)^{2017}.\left(-2017\right)^{2018}\)
\(A=0.\left(-2016\right)^{2017}.\left(-2017\right)^{2018}\)
\(A=0\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{2015}{2016}+\frac{2016}{2017}>\frac{\left(2015+2016\right)}{\left(2016+2017\right)}=\frac{2015}{2016+2017}+\frac{2016}{2016+2017}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=\frac{2015}{2016}+\frac{2016}{2017}=1-\frac{1}{2016}+1-\frac{1}{2017}>1\)
\(B=\frac{2015+2016}{2016+2017}< \frac{2016+2017}{2016+2017}=1\)
Suy ra \(A>B\).
![](https://rs.olm.vn/images/avt/0.png?1311)
a) 20162=2016.2016=2016.(2015+1)=2016.2015+2016
2015.2017=2015.(2016+1)=2015.2016+2015
Vì 2016>2015 => 2016.2015+2016>2015.2016+2015
=>20162>2015.2017
b) 35.53-18=(34+1).53-18=34.53-53-18=34.53+35
Vì 34.53+35=35+53.34 => 35.53-18=35+53.34
a) 20162=2016.2016=2016.(2015+1)=2016.2015+2016
2015.2017=2015.(2016+1)=2015.2016+2015
Vì 2016>2015 => 2016.2015+2016>2015.2016+2015
=>20162>2015.2017
b) 35.53-18=(34+1).53-18=34.53-53-18=34.53+35
Vì 34.53+35=35+53.34 => 35.53-18=35+53.34
Ai trên 10 điểm hỏi đáp thì mình nha mình đang cần gấp chỉ còn 51 điểm là tròn rồi mong các bạn hỗ trợ mình sẽ đền bù xứng đáng
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có :
\(\frac{2013}{2014}< 1\)( 1 )
\(\frac{2014}{2015}< 1\)( 2 )
\(\frac{2015}{2016}< 1\)( 3 )
từ ( 1 ) , ( 2 ) và ( 3 )
\(\Rightarrow\frac{2013}{2014}+\frac{2014}{2015}+\frac{2015}{2016}< 1+1+1=3\)
vậy A < 3
Có: 2013/2014<2014/2014
2014/2015<2015/2015
2015/2016<2016/2016
=>2013/2014+2014/2015+2015/2016<2014/2014+2015/015+2016/2016
=>A<3
ta có 2015/2016+2016/2017+2017/2015=(1-1/2016)+(1-1/2017)+(2+1/2015)
=4-(1/2016+1/2017-1/2015)
1/2016<1; 1/2017<1 nên 1/2016+1/2017<2 suy ra 1/2016+1/2017-1/2015<1(vì 1/2015<1)
4-(1/2016+1/2017-1/2015)>4-1=3
2015/2016+2016/2017+2017/2015>3
cho mik nhé