Question

Answer 1

Lời giải:

a)

\(A=\frac{10^8+2}{10^8-1}=1+\frac{3}{10^8-1}< 1+\frac{3}{10^8-3}=\frac{10^8}{10^8-3}=B\)

Vậy $A<B$

b)

 \(17C=\frac{17^{204}+17}{17^{204}+1}=1+\frac{16}{17^{204}+1}\)

\(17D=\frac{17^{203}+17}{17^{203}+1}=1+\frac{16}{17^{203}+1}\)

Mà \(\frac{16}{17^{204}+1}< \frac{16}{17^{203}+1}\Rightarrow 17C< 17D\Rightarrow C< D\)

c) 

\(M=\frac{13(13^{18}+1)-12}{13^{18}+1}=13-\frac{12}{13^{18}+1}\)

\(N=\frac{13(13^{19}+1)-12}{13^{19}+1}=13-\frac{12}{13^{19}+1}\)

Mà \(\frac{12}{13^{18}+1}>\frac{12}{13^{19}+1}\Rightarrow M< N\)