a) \(\frac{1995}{1997}\)và \(\frac{1995}{1996}\)

Ta có : \(\frac{1995}{1996}=\frac{1995\times2}{1996\times2}=\frac{3990}{3992}\)

\(1-\frac{1995}{1997}=\frac{2}{1997};1-\frac{3990}{3992}=\frac{2}{3992}\)

Vì \(\frac{2}{1997}>\frac{2}{3992}\)nên \(\frac{1995}{1997}< \frac{3990}{3992}\)hay \(\frac{1995}{1997}< \frac{1995}{1996}\).

b) \(\frac{2016}{2017}\)và \(\frac{2017}{2018}\)

Ta có : \(1-\frac{2016}{2017}=\frac{1}{2017};1-\frac{2017}{2018}=\frac{1}{2018}\)

Vì \(\frac{1}{2017}>\frac{1}{2018}\)nên \(\frac{2016}{2017}< \frac{2017}{2018}\).

c) \(\frac{2018}{2019}\)và \(\frac{2017}{2016}\).

Vì \(\frac{2018}{2019}< 1;1< \frac{2017}{2016}\)nên \(\frac{2018}{2019}< \frac{2017}{2016}\).

~ HOK TỐT ~

a)\(\frac{1995}{1997}\)<   \(\frac{1995}{1996}\)

b)\(\frac{2016}{2017}\)<    \(\frac{2017}{2018}\)

c)\(\frac{2018}{2019}\)<     \(\frac{2017}{2016}\)

\(a,\frac{1995}{1997}< \frac{1995}{1996}\)

\(b,\frac{2016}{2017}< \frac{2017}{1208}\)

\(c,\frac{2018}{2019}< \frac{2017}{2016}\)

A>B do A>4 cònB<4

hình như cái này đâu phải toán lớp 5 đâu bạn

nhầm toán lớp 6

12+13×14

 mình với

á/ 2017 mù 2018 = chữ số tận cùng là 30

2017\(^{2018}\)=  (2017\(^2\))\(^{1009}\)= ( .....9)\(^{1009}\)= (....9)

2016\(^{2017}\)+ 2017\(^{2018}\)+ 2018\(^{2019}\)

=  (....6) + (...9) + (...6)

=(...1)

\(A=\frac{2016}{2017}+\frac{2017}{2018}+\frac{2018}{2019}\)

\(\Rightarrow A=(1-\frac{1}{2017})+(1-\frac{1}{2018})+(1-\frac{1}{2019})\)

\(\Rightarrow A=3-\left(\frac{1}{2017}+\frac{1}{2018}+\frac{1}{2019}\right)\)

\(\left(\frac{1}{2017}+\frac{1}{2018}+\frac{1}{2019}\right)\)<\(\frac{3}{2017}\)<\(1\)

\(\Rightarrow A\)>\(3-1=2\)

\(B=\frac{2016+2017+2018}{2017+2018+2019}\)

\(\Rightarrow B=1-\frac{3}{6054}\)

\(\Rightarrow B=1-\frac{1}{2018}\)

\(B\)<\(1\);\(A\)>\(2\)

\(\Rightarrow A\)>\(B\)