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....

ai trả lời đúng mik tick cho

\(A=\dfrac{3^{123}+1}{3^{125}+1}\Leftrightarrow3^2A=\dfrac{3^{125}+9}{3^{125}+1}\)

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

\(B=\dfrac{3^{122}+1}{3^{124}+1}\Leftrightarrow3^2B=\dfrac{3^{124}+9}{3^{124}+1}\)

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

\(9A< 9B\Leftrightarrow A< B\)

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

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

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

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

Nên A<B

bạn giải dùm mik bài đó luk được hok

\(\frac{x+1}{125}+\frac{x+2}{124}+\frac{x+3}{123}+\frac{x+4}{122}+\frac{x+146}{5}=0\)

\(\left(\frac{x+1}{125}+1\right)+\left(\frac{x+2}{124}+1\right)+\left(\frac{x+3}{123}+1\right)+\left(\frac{x+4}{122}+1\right)+\left(\frac{x+146}{5}-4\right)=0\)

\(\frac{x+126}{125}+\frac{x+126}{124}+\frac{x+126}{123}+\frac{x+126}{122}+\frac{x+126}{5}=0\)

\(\left(x+126\right).\left(\frac{1}{125}+\frac{1}{124}+\frac{1}{123}+\frac{1}{122}+\frac{1}{5}\right)=0\)

vì \(\left(\frac{1}{125}+\frac{1}{124}+\frac{1}{123}+\frac{1}{122}+\frac{1}{5}\right)\ne0\)nên x + 126 = 0 \(\Rightarrow\)x = -126