Question

[Phan Quang Minh]

so sánh A = 2016¹⁰+1/2016¹¹+1 và 2016¹¹+1/2016¹²+1    thong cam ko biet viet tu so mau so

Answer 1

[Nguyễn Hưng Phát]

Ta có:2016A=\(\frac{2016\left(2016^{10}+1\right)}{2016^{11}+1}=\frac{2016^{11}+2016}{2016^{11}+1}=\frac{2016^{11}+1+2015}{2016^{11}+1}=1+\frac{2015}{2016^{11}+1}\)

2016B=\(\frac{2016\left(2016^{11}+1\right)}{2016^{12}+1}=\frac{2016^{12}+2016}{2016^{12}+1}=\frac{2016^{12}+1+2015}{2016^{12}+1}=1+\frac{2015}{2016^{12}+1}\)

Vì \(2016^{12}+1>2016^{11}+1\) nên \(\frac{2015}{2016^{12}+1}< \frac{2015}{2016^{11}+1}\)

\(\Rightarrow1+\frac{2015}{2016^{12}+1}< 1+\frac{2015}{2016^{11}+1}\)\(\Rightarrow2016B< 2016A\Rightarrow B< A\)

Answer 2

[Thanh Hằng Nguyễn]

Ta có :

\(B=\frac{2016^{11}+1}{2016^{12}+1}>\frac{2016^{11}+1+9}{2016^{12}+1+9}=\frac{2016^{11}+10}{2016^{12}+10}=\frac{10\left(2016^{10}+1\right)}{10\left(2016^{11}+1\right)}=\frac{2016^{10}+1}{2016^{11}+1}=A\)

Vậy \(A< B\)