Question

[Hoàng Thị Ngọc Trâm]

So sánh:

A= 17 mũ 18 + 1 trên 17 mũa 19 +1

B= 17 mũ 17 + 1 trên 17 mũ 18 + 1

giúp tớ nha

Answer 1

[Nguyễn Linh Nhi]

Ta có:

$A=\frac{{17}^{18}+1}{{17}^{19}+1}$

$\Rightarrow 17A=\frac{{17}^{19}+1+16}{{17}^{19}+1}$

$\Rightarrow 17A=1+\frac{16}{{17}^{19}+1}$

$B=\frac{{17}^{17}+1}{{17}^{18}+1}$

$\Rightarrow 17B=\frac{{17}^{18}+1+16}{{17}^{18}+1}$

$\Rightarrow 17B=1+\frac{16}{{17}^{18}+1}$

Vì $\frac{16}{{17}^{19}+1}<\frac{16}{{17}^{18}+1}\Rightarrow 17A<17B$

$\Rightarrow A<B$

Vậy $A<$$B$