7278,085

43/47>19/35

43/49<31/35

\(\frac{29}{35}\)<\(\frac{43}{49}\)<\(\frac{31}{35}\)<\(\frac{43}{47}\)

43/47=43x35/47x35=1505/1645

                                                  => 1505/1645>1363/1645 <=> 43/47>29/35

29/35=29x47/35x47=1363/1645

43/49=43x35/49x35=1505/1715

                                                   => 1505/1715<1519/1715 <=> 43/49<31/35

31/35=31x49/35x49=1519/1715

\(\frac{3}{111}< \frac{4}{111}< \frac{4}{103}\)

\(\frac{91}{93}=\frac{93-2}{93}=1-\frac{2}{93}\)

\(\frac{105}{107}=\frac{107-2}{107}=1-\frac{2}{107}\)

\(\Rightarrow\frac{91}{93}< \frac{105}{107}\)

\(\frac{71}{35}=\frac{70+1}{35}=2+\frac{1}{35}=2+\frac{2}{70}\)

\(\frac{127}{53}=\frac{106+11}{53}=2+\frac{11}{53}\)

\(\frac{2}{70}< \frac{2}{53}< \frac{11}{53}\Rightarrow\frac{71}{35}< \frac{127}{53}\)

Ta có : \(2a=3b\Rightarrow\dfrac{a}{3}=\dfrac{b}{2};5b=7c\Rightarrow\dfrac{b}{7}=\dfrac{c}{5}\)

\(\Rightarrow\dfrac{a}{21}=\dfrac{b}{14}=\dfrac{c}{10}\)

Áp dụng tính chất dãy tỉ số bằng nhau ta có : 

\(\dfrac{a}{21}=\dfrac{b}{14}=\dfrac{c}{10}=\dfrac{3a-7b+5c}{3.21-7.14+5.10}=\dfrac{76}{15}\)

\(a=\dfrac{532}{5};b=\dfrac{7448}{5};c=14896\)

\(2^{91}=\left(2^{13}\right)^7=8192^7\)

\(5^{35}=\left(5^5\right)^7=3125^7\)

Ta có 8192 > 3125

Nên \(8125^7>3125^7\)

Vậy : \(2^{91}>5^{35}\)

Ta có
\(2^{91}=\left(2^{13}\right)^7=8192^7\)

\(5^{35}=\left(5^5\right)^7=3125^7\)

Vì \(8192^7>3125^7\) ( 8192 > 3125 ) nên \(2^{91}>5^{35}\)

Vậy \(2^{91}>5^{35}\)

2⁹¹ > 5³⁵

A = 35.53-18

=(34+1).53-18

=34.53+35

=> A=B

A=35.53-18

=(34+1).53-18

=34.53+53-18

=34.53+35

=>A=B