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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
\(A=\dfrac{3^{123}+1}{3^{125}+1}\Leftrightarrow3^2A=\dfrac{3^{125}+9}{3^{125}+1}\)
\(9A=\dfrac{3^{125}+1}{3^{125}+1}+\dfrac{8}{3^{125}+1}=1+\dfrac{8}{3^{125}+1}\)
\(B=\dfrac{3^{122}+1}{3^{124}+1}\Leftrightarrow3^2B=\dfrac{3^{124}+9}{3^{124}+1}\)
\(9B=\dfrac{3^{124}+1+8}{3^{124}+1}+\dfrac{3^{124}+1}{3^{124}+1}+\dfrac{8}{3^{124}+1}=1+\dfrac{8}{3^{124}+1}\)
\(9A< 9B\Leftrightarrow A< B\)
Đề đúng là \(B=\frac{3^{122}+1}{3^{124}+1}\)nhé .
Ta có :
\(9A=9.\left(\frac{3^{123}+1}{3^{125}+1}\right)=\frac{3^{125}+9}{3^{125}+1}\)
\(=1+\frac{8}{3^{125}+1}\)
\(9B=9.\left(\frac{3^{122}+1}{3^{124}+1}\right)=\frac{3^{124}+9}{3^{124}+1}\)
\(=1+\frac{8}{3^{124}+1}\)
Dễ thấy \(3^{124}+1< 3^{125}+1\)
\(\Leftrightarrow\frac{8}{3^{125}+1}< \frac{8}{3^{124}+1}\)
\(\Leftrightarrow\frac{8}{3^{125}+1}+1< \frac{8}{3^{124}+1}+1\)
\(\Leftrightarrow A< B\)
Vậy....
\(A=1+3+3^2+3^3+...+3^{2016}\)
\(A=1+3\left(1+3^2+...+3^{2015}\right)\)
\(A=1+3\left(A-3^{2016}\right)\)
\(A=1+3A-3^{2017}\)
\(2A=3^{2017}-1\Rightarrow A=\frac{3^{2017}-1}{2}\)
\(A< B\)
ai trả lời đúng mik tick cho