Question

So sánh A và B, biết:

A= 2018/2019 + 2019/2018 

và B= 2

Trả lời nhanh giúp mk nhé, mk đang cần gấp!

Answer 1

A nhỏ hơn B

Answer 2

Ta có :

\(\frac{2018}{2019}\)\(+\)\(\frac{2019}{2018}\)\(=\frac{2018}{2019}\)\(+\frac{1}{2018}\)\(+1>\frac{2018+1}{2019}\)\(+1\)

\(=1+1=2\)

\(\Rightarrow\frac{2018}{2019}\)\(+\frac{2019}{2018}\)\(>2\)

\(\Rightarrow A>B\)

Answer 3

Xét \(A=\frac{2018}{2019}+\frac{2019}{2018}=\left(1-\frac{1}{2019}\right)+\left(1+\frac{1}{2018}\right)\)

\(=1-\frac{1}{2019}+1+\frac{1}{2018}=\left(1+1\right)+\left(\frac{1}{2018}-\frac{1}{2019}\right)\)

\(=2+\left(\frac{1}{2018}-\frac{1}{2019}\right)\). Mà \(\frac{1}{2018}>\frac{1}{2019}\Rightarrow2+\left(\frac{1}{2018}-\frac{1}{2019}\right)>2\Leftrightarrow A>B\)

Answer 4

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

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

\(A=2-\frac{1}{2019}+\frac{1}{2018}\)

\(\text{Ta có: }\frac{1}{2019}< \frac{1}{2018}\Rightarrow-\frac{1}{2019}+\frac{1}{2018}>0\)

\(\Rightarrow2-\frac{1}{2019}+\frac{1}{2018}>2\)

\(\Rightarrow A>B\)