Question

so sánh A và B:

A=\(\frac{3^{123}+1}{3^{125}+1}\)

B=\(\frac{3^{122}+1}{3^{124}+1}\)

Answer 1

Áp dụng \(\frac{a}{b}< 1\Leftrightarrow\frac{a}{b}< \frac{a+m}{b+m}\) (a;b;m \(\in\)N*)

Ta có:

\(A=\frac{3^{123}+1}{3^{125}+1}< \frac{3^{123}+1+2}{3^{125}+1+2}\)

\(A< \frac{3^{123}+3}{3^{125}+3}\)

\(A< \frac{3.\left(3^{122}+1\right)}{3.\left(3^{124}+1\right)}\)

\(A< \frac{3^{122}+1}{3^{124}+1}=B\)

=> A < B

Answer 2

\(9A=\frac{3^{125}+9}{3^{125}+1}=1+\frac{8}{3^{125}+1}\)

\(9B=\frac{3^{124}+9}{3^{124}+1}=1+\frac{8}{3^{124}+1}\)

Mà 3^125+1>3^124+1         =>\(\frac{8}{3^{125}+1}< \frac{8}{3^{124}+1}\)

Nên A<B

Answer 3

9A=\(\frac{3^{125}+9}{3^{125}+1}\)=\(1+\frac{8}{3^{125}+1}\)

9B=\(\frac{3^{124}+9}{3^{124}+1}\)=\(1+\frac{8}{3^{124}+1}\)

Vì \(\frac{8}{3^{125}+1}< \frac{8}{3^{124}+1}\)\(\Rightarrow9B>9A\)\(\Rightarrow B>A\)

Vậy B>A