Ta có : \(\frac{-2525}{2929}=\left(-1\right)\frac{25}{29}\)

\(\frac{-217}{245}=\left(-1\right)\frac{217}{245}\)

Xét \(\frac{25}{29}\)và \(\frac{217}{245}\)

Có 25 . 245 = 6125

217 . 29 = 6293

Có 6293>6125

=> 25/29<217/245

=>-25/29>-217/245

2                                                                

2                            

2 nha bạn

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 \(\frac{27}{82}\)<\(\frac{26}{75}\)