Question

So sánh A=\(\frac{2020^{10}+2}{2020^{11}+2}\)và B=\(\frac{2020^{11}+2}{2020^{12}+2}\)

Answer 1

ta có \(A=\frac{2020^{10}+2}{2020^{11}+2}=>2020A=\frac{2020^{11}+4040}{2020^{11}+2}=1+\frac{4038}{2020^{11}+2}\)(1)

\(B=\frac{2020^{11}+2}{2020^{12}+2}=>2020B=\frac{2020^{12}+4040}{2020^{12}+2}=1+\frac{4038}{2012^{12}+2}\)(2)

từ 1 and 2 => 2020B<2020A

=> A>B

Answer 2

Ta có B=\(\frac{2020^{11}+2}{2020^{12}+2}\)

suy ra \(B< \frac{\left(2020^{11}+2\right)+2018}{\left(2020^{12}+2\right)+2018}=\frac{2020^{11}+2020}{2020^{12}+2020}=\frac{2020\left(2020^{10}+2\right)}{2020\left(2020^{11}+2\right)}=\frac{2020^{10}+2}{2020^{11}+2}\)

nên A > B