\(3^{203}=3^{200}.3^3=9^{100}.27\)

\(2^{302}=2^{300}.2^2=8^{100}.4\)

Vì \(9^{100}>8^{100}\); \(27>4\)\(\Rightarrow3^{203}>2^{302}\)

\(3^{203}>3^{202}=\left(3^2\right)^{101}=9^{101}\)

\(2^{302}< 2^{303}=\left(2^3\right)^{101}=8^{101}\)

\(\Rightarrow2^{302}< 8^{101}< 9^{101}< 3^{203}\)

\(3^{203}>2^{302}\)

hok tốt

#sakurasyaoran#

Dấu > nha !

\(3^{203}\) và \(2^{302}\)

Ta có:

\(3^{203}>3^{202}=\left(3^2\right)^{101}=9^{101}.\)

\(2^{302}< 2^{303}=\left(2^3\right)^{101}=8^{101}.\)

Vì \(9>8\) nên \(9^{101}>8^{101}.\)

\(\Rightarrow3^{203}>2^{302}.\)

Chúc bạn học tốt!

thankss nhaa

a) \(\dfrac{11}{2}=\dfrac{10+1}{2}=5+\dfrac{1}{2}\)

\(\dfrac{32}{9}=\dfrac{27+5}{9}=3+\dfrac{5}{9}< 5+\dfrac{1}{2}\)

Vậy \(\dfrac{11}{2}>\dfrac{32}{9}\)

b)\(\dfrac{100}{23}=\dfrac{92+8}{23}=4+\dfrac{8}{23}\)

 \(\dfrac{302}{123}=\dfrac{246+56}{123}=2+\dfrac{56}{123}< 4+\dfrac{8}{23}\)

Vậy \(\dfrac{100}{23}>\dfrac{302}{123}\)

c) \(\dfrac{515}{605}< \dfrac{515+1}{605+1}=\dfrac{516}{606}\Rightarrow\dfrac{515}{605}< \dfrac{516}{606}\)

 * 5^302 = 25.5^300 = 25.(5^3)^100 = 25.125^100 
11^201= 11.11^200 = 11.(11^2)^100 = 11.121^100 
125^100 > 121^100 Vậy 5^302 > 11^201