A = \(\frac{2013}{2014}+\frac{2014}{2015}>\frac{1}{2}+\frac{1}{2}=1\)

\(B=\frac{2013+2014+2015}{2014+2015+2016}<1\)

\(Vậy:A>B\)

Đúng nha Nguyễn Bình Minh

so sánh:

\(A=\frac{2013}{2014}+\frac{2014}{2015}+\frac{2015}{2016}\)  và\(B=\) \(\frac{2013+2014+2015}{2014+2015+2016}\)

                                                             \(B=\frac{2013}{2014+2015+2016}+\frac{2014}{2014+2015+2016}+\frac{2015}{2014+2015+2016}\)

Ta có: \(\frac{2013}{2014}>\frac{2013}{2014+2015+2016}\)

          \(\frac{2014}{2015}>\frac{2014}{2014+2015+2016}\)

          \(\frac{2015}{2016}>\frac{2015}{2014+2015+2016}\)

\(\Rightarrow\frac{2013}{2014}+\frac{2014}{2015}+\frac{2015}{2016}>\frac{2013+2014+2015}{2014+2015+2016}\)

Vậy: \(A>B\)

bằng nhau nha

a)\(\frac{2013}{2015}< \frac{2014}{2016}\)

b)\(\frac{2013+2014}{2014+2015}< \frac{2013}{2014}+\frac{2014}{2015}\)

ta có tính chất \(\frac{a}{b}\)>1 suy ra \(\frac{a.m}{b.m}\).........

A=1-1/(2013*2014)

B=1-1/(2014*2015)

2013*2014<2014*2015

=>1/2013*2014>1/2014*2015

=>-1/2013*2014<-1/2014*2015

=>A<B

\(C=\dfrac{2013}{2013}+2014+\dfrac{2014}{2014}+2015+\dfrac{2015}{2015}+2016\)

\(=1+2014+1+2015+1+2016\)

\(=6048>2\)

Vậy: \(C>D\)

Bạn à, đây không phải là toán lớp 5 nên mình không giải được nên bạn thông cảm nha!

UK ĐÂY TOÁN 6 ĐÓ

M < N

Đáp án B