a, >

b, <

c, <

cái nào mũ lớn hơn thì nó lớn lơn

a)Ta có : \(3^{39}=\left(3^{13}\right)^3=1594323^3\)
  \(11^{21}=\left(11^7\right)^3=19487171^3\)
Vì \(1594323< 19487171\)
\(=>1594323^3< 19487171^3\)
\(=>3^{39}< 11^{21}\)
Vậy \(3^{39}< 11^{21}\)
   b)Ta có : \(199^{20}=\left(199^4\right)^5=1568239201^5\)
                \(2002^{15}=\left(2002^3\right)^5=8024024008^5\)
Vì \(1568239201< 8024024008\)
\(=>1568239201^5< 8024024008^5\)
\(=>199^{20}< 2002^{15}\)
Vậy \(199^{20}< 2002^{15}\)
c) Ta có:\(125^{90}=\left(125^3\right)^{30}=1953125^{30}\)
            \(25^{120}=\left(25^4\right)^{30}=390625^{30}\)
Vì \(1953125>390625\)
\(=>1953125^{30}>390625^{30}\)
\(=>125^{90}>25^{120}\)
Vậy \(125^{90}>25^{120}\)
d)Ta có : \(3^{500}=\left(3^5\right)^{100}=243^{100}\)
              \(7^{300}=\left(7^3\right)^{100}=343^{100}\)
Vì \(243< 343\)
\(=>243^{100}< 343^{100}\)
\(=>3^{500}< 7^{300}\)
Vậy \(3^{500}< 7^{300}\)
Phù , cuối cùng cũng viết xong . Mỏi tay quá ! À , chúc bạn học tốt nhé !

\(\)

-25/20 < 20/25

15/21 > 21/49

ᵃ, ᵗᵃ ʳᵘ́ᵗ ᵍᵒ̣ⁿ :\(\dfrac{18}{24}=\dfrac{3}{4}\) ;

 ᵗᵃ ʳᵘ́ᵗ ᵍᵒ̣ⁿ :\(\dfrac{15}{25}=\dfrac{3}{5}\)

Vì \(\dfrac{3}{4}>\dfrac{3}{5}\\ \Rightarrow\dfrac{18}{24}>\dfrac{15}{25}\)

ᵇ, ᵗᵃ ʳᵘ́ᵗ ᵍᵒ̣ⁿ :\(\dfrac{21}{15}=\dfrac{7}{5}\\ \dfrac{28}{24}=\dfrac{7}{6}\)

Vì \(\dfrac{7}{5}>\dfrac{7}{6}\\ \Rightarrow\dfrac{21}{15}>\dfrac{28}{24}\)

a, \(\dfrac{18}{24}\) = \(\dfrac{3}{4}\)                            \(\dfrac{15}{25}\) = \(\dfrac{3}{5}\) 

     \(\dfrac{3}{4}\)  >  \(\dfrac{3}{5}\) vậy \(\dfrac{18}{24}\) > \(\dfrac{15}{25}\)

b, \(\dfrac{21}{15}\) = \(\dfrac{7}{5}\)                            \(\dfrac{28}{24}\) = \(\dfrac{7}{6}\)

      \(\dfrac{7}{5}\)  >  \(\dfrac{7}{6}\) vậy \(\dfrac{21}{15}\)  >  \(\dfrac{28}{24}\) 

21/35=3/5; 12/20=3/5

=>21/35=12/20

24/35<1<21/7

35/49=5/7; 40/64=5/8

=>35/49>40/64

2424/4848=1/2

9/11=1-2/11

13/15=1-2/15

mà 2/11>2/15

nên 9/11<13/15

19/15>11/15

khó

a>\(16^{19}=2^{4^{19}}=2^{76}\\ 8^{25}=2^{3^{25}}=2^{75}\)

16^19>8^25

a:Cách 1: \(\frac{21}{35}=\frac{21:7}{35:7}=\frac35\)

\(\frac{12}{40}=\frac{12:4}{40:4}=\frac{3}{10}\)

mà \(\frac35>\frac{3}{10}\left(5<10\right)\)

nên \(\frac{21}{35}>\frac{12}{40}\)

Cách 2: \(\frac{21}{35}=\frac{21\times8}{35\times8}=\frac{168}{280}\)

\(\frac{12}{40}=\frac{12\times7}{40\times7}=\frac{84}{280}\)

mà 168>84

nên \(\frac{21}{35}>\frac{12}{40}\)

b: Cách 1: \(\frac{35}{49}=\frac{35:7}{49:7}=\frac57\)

\(\frac{40}{64}=\frac{40:8}{64:8}=\frac58\)

mà \(\frac57>\frac58\left(7<8\right)\)

nên \(\frac{35}{49}>\frac{40}{64}\)

Cách 2: \(\frac{35}{49}=\frac{35\times64}{49\times64}=\frac{2240}{3136}\)

\(\frac{40}{64}=\frac{40\times49}{64\times49}=\frac{1960}{3136}\)

mà 2240>1960

nên \(\frac{35}{49}>\frac{40}{64}\)

c: Cách 1: \(\frac{9}{11}=1-\frac{2}{11}\)

\(\frac{13}{15}=1-\frac{2}{15}\)

Ta có: 11<15

=>\(\frac{2}{11}>\frac{2}{15}\)

=>\(-\frac{2}{11}<-\frac{2}{15}\)

=>\(-\frac{2}{11}+1<-\frac{2}{15}+1\)

=>\(\frac{9}{11}<\frac{13}{15}\)

Cách 2: Ta có: \(\frac{9}{11}=\frac{9\times15}{11\times15}=\frac{135}{165}\)

\(\frac{13}{15}=\frac{13\times11}{15\times11}=\frac{143}{165}\)

mà 135<143

nên \(\frac{9}{11}<\frac{13}{15}\)

d: Cách 1: \(\frac{21}{7}=\frac{21\times5}{7\times5}=\frac{105}{35}\)

mà 105>24

nên \(\frac{21}{7}>\frac{24}{35}\)

cách 2: Ta có: \(\frac{21}{7}=3>1;1>\frac{24}{35}\)

Do đó: \(\frac{21}{7}>\frac{24}{35}\)

e: Cách 1: \(\frac12=\frac{1\times2424}{2\times2424}=\frac{2424}{4848}\)

Cách 2: \(\frac{2424}{4848}=\frac{2424:2424}{4848:2424}=\frac12\)

f: Vì 19>11

nên \(\frac{19}{15}>\frac{11}{15}\)