A=\(\frac{3^{10}+1}{3^9+1}\)>1
=> A=\(\frac{3^{10}+1}{3^9+1}\)> \(\frac{3^{10}+1+2}{3^9+1+2}\)
=>A=\(\frac{3^{10}+1}{3^9+1}\)>\(\frac{3^{10}+3}{3^9+3}\)
=>A=\(\frac{3^{10}+1}{3^9+1}\)>\(\frac{3\left(3^9+1\right)}{3.\left(3^8+1\right)}\)
=>A=\(\frac{3^{10}+1}{3^9+1}\)>\(\frac{3^9+1}{3^8+1}\)=B
vậy A>B
vì 1 là bạn quy đồng lên
hai là giải thich như sau
\(\frac{25}{29}\)= 1- \(\frac{4}{29}\)
\(\frac{31}{35}\)= 1- \(\frac{4}{35}\) mà \(\frac{4}{29}\)> \(\frac{4}{35}\)
nên 1-\(\frac{4}{29}\)<1-\(\frac{4}{35}\)
vậy \(\frac{25}{29}\)<\(\frac{31}{35}\)
a. \(\frac{-2525}{2929}\)= -\(\frac{25.101}{29.101}\)=\(\frac{-25}{29}\)
\(\frac{-217}{245}\)=\(\frac{-7.31}{7.35}\)=\(\frac{-31}{35}\)
vì \(\frac{25}{29}\)>\(\frac{31}{35}\) => \(\frac{-25}{29}\)<\(\frac{-31}{35}\)
b.ta có 27.75=2025
26.82=2132
=> 26.82> 27.75
vậy \(\frac{27}{82}\)<\(\frac{26}{75}\)
Mai Linh cho mình hỏi bạn lm cách nào mà ra đc 25/29>31/35 z ?
à quên \(\frac{25}{29}\)<\(\frac{31}{35}\)=> \(\frac{-25}{29}\)>\(\frac{-31}{35}\)