Question

So sánh 

\(A=\frac{2015}{2016}+\frac{2016}{2017}\) và \(B=\frac{2015+2016}{2016+2017}\)

Giải cả bài nha

Answer 1

dễ thấy B=\(\frac{2015+2016}{2016+2017}\)<1

A=\(\frac{2015}{2016}\)+\(\frac{2016}{2017}\)=1-\(\frac{1}{2016}\)+1-\(\frac{1}{2017}\)=(1+1)-(\(\frac{1}{2016}\)+\(\frac{1}{2017}\))=2-(\(\frac{1}{2016}\)+\(\frac{1}{2017}\))

vì (\(\frac{1}{2016}\)+\(\frac{1}{2017}\))<0,5+0,5=1 suy ra 2-(\(\frac{1}{2016}\)+\(\frac{1}{2017}\))>1 mà b<1suy ra A>B

 

Answer 2

Ta thấy: B=\(\frac{2015+2016}{2016+2017}\)=\(\frac{2015}{2016+2017}\)+\(\frac{2016}{2016+2017}\)

              A=\(\frac{2015}{2016}\)+\(\frac{2016}{2017}\)

Mà\(\frac{2015}{2016+2017}\)<\(\frac{2015}{2016}\); \(\frac{2016}{2016+2017}\)<\(\frac{2016}{2017}\)

Suy ra: \(\frac{2015}{2016}\)+\(\frac{2016}{2017}\)>\(\frac{2015}{2016+2017}\)+\(\frac{2016}{2016+2017}\)=\(\frac{2015+2016}{2016+2017}\)

               Hay A>B

Answer 3

Tại sao nói nghành ruột khoang chủ yếu có lợi

Answer 4

dễ thấy B=\(\dfrac{2015+2016}{2016+2017}\)<1

A=\(\dfrac{2015}{2016}\)+\(\dfrac{2016}{2017}\)=1-\(\dfrac{1}{2016}\)+1-\(\dfrac{1}{2017}\)=(1+1)-(\(\dfrac{1}{2016} \)+\(\dfrac{1}{2017}\))

vì (\(\dfrac{1}{2016}\)+\(\dfrac{1}{2017}\))<0,5+0,5=1 suy ra 2-(\(\dfrac{1}{2016}\)+\(\dfrac{1}{2017}\))>1 mà b<1 suy ra A>B