cứu tui 

Ta có:\(2^{30}=\left(2^3\right)^{10}=8^{10}< 9^{10}=\left(3^2\right)^{10}=3^{20}\)

\(3^{30}=3^{20}.3^{10}< 3^{20}.4^{10}=3^{20}.\left(2^2\right)^{10}=3^{20}.2^{20}=\left(3.2\right)^{20}=6^{20}\)

\(4^{30}=4^{20}.4^{10}=4^{20}.\left(2^2\right)^{10}=4^{20}.2^{20}=\left(4.2\right)^{20}=8^{20}\)

nên \(2^{30}+3^{30}+4^{30}< 3^{20}+6^{20}+8^{20}\)

Ta có : x=2³⁰ + 3³⁰ + 4³⁰

x=2^3.10 + 3^3.10 + 4^3.10

x=(2³)²⁰ + (3³)¹⁰ + (4³)¹⁰

x=8²⁰ + 9¹⁰ + 64¹⁰

y=3²⁰ + 6²⁰ + 8²⁰

y=3^2.10 + 6^2.10 + 8^2.10

y=(3²)¹⁰ + (6²)¹⁰ + (8²)¹⁰

y=9¹⁰ + 36¹⁰ + 64¹⁰

mà 9¹⁰ > 8²⁰ ;36¹⁰ >9¹⁰ ;64¹⁰ = 64¹⁰

nên x<y

a)\(10^{20}=\left(10^2\right)^{10}=100^{10}\left(1\right)\)

\(9^{30}=\left(9^3\right)^{10}=729^{10}\left(2\right)\)

Từ (1) và (2) \(\Rightarrow9^{30}>10^{20}\)

b) \(\left(-5\right)^{30}=5^{30}=125^{10}\)

\(\left(-3\right)^{50}=3^{50}=243^{10}\)

\(\Rightarrow\left(-3\right)^{50}>\left(-5\right)^{30}\)

c)\(64^8=\left(2^6\right)^8=2^{48}\)

\(16^{12}=\left(2^4\right)^{12}=2^{48}\)

\(\Rightarrow64^8=16^{12}\)

Xét \(A=2^{30}+3^{30}+4^{30}=\left(2^3\right)^{10}+\left(3^3\right)^{10}+\left(2^2\right)^{30}=8^{10}+27^{10}+2^{60}\)

\(B=3^{20}+6^{20}+8^{20}=\left(3^2\right)^{10}+\left(6^2\right)^{10}+\left(2^3\right)^{20}=9^{10}+36^{10}+2^{60}\)

Vì \(8^{10}< 9^{10},27^{10}< 36^{10}\)nên A<B

2³⁰ = 2^3.10= 8¹⁰

3³⁰=3^3.10=27¹⁰

4³⁰=4^3.10=64¹⁰

Vế trái = 8¹⁰ + 27¹⁰ + 64¹⁰

3²⁰=3^2.10=9¹⁰

6²⁰=6^2.10=36¹⁰

8²⁰=8^2.10=64¹⁰

vế phải = 9¹⁰ + 36¹⁰ + 64¹⁰

Vì 64¹⁰=64¹⁰ ; 36¹⁰ > 27¹⁰ ; 9¹⁰ > 8¹⁰

=> vế phải > vế trái

so sánh: 2^30 + 3^30 + 4^30 và 3^20 + 6^20 + 8^20
2^30 = ( 2^3)^10 = 8^ 10
3^30 = (3^3)^10 = 27^10
4^30 = (4^3)^10 = 64^10
3^20 = (3^2)^10 = 9^10
6^20 = (6^2) = 36^10
8^20 = (8^2)^10 = 84^10
vì 9^10 > 8^10
36^10 > 27^10
84^10 > 64^10
=> 2^30 + 3^30 + 4^30 < 3^20 + 6^20 + 8^20

a) \(A=1+2+2^2+2^3+...+2^{100}\) \(B=2^{201}\)

\(2A=2\left(1+2+2^2+2^3+...+2^{100}\right)\)

\(2A=2+2^2+2^3+2^4+...+2^{201}\)

\(2A-A=\left(2+2^2+2^3+2^4+...+2^{201}\right)-\left(1+2+2^2+2^3+...+2^{100}\right)\)

\(2A-A=2^{101}-1\)

\(A=2^{201}-1\)

Ta có 2²⁰¹ > 2²⁰¹ - 1 => B > A => 2²⁰¹ > 1 + 2 + 2² + 2³ +...+ 1¹⁰⁰

b) 2¹⁰⁰ = 2³¹ . 2⁶³ . 2⁶ = 2³¹ . (2⁹)⁷ . (2²)³ = 2³¹ . 512⁷ . 4³ (1)

10³¹ = 2³¹ . 5²⁸ . 5³ = 2³¹ . (5⁴)⁷ . 5³ = 2³¹ . 625⁷ . 5³ (2)

Từ (1) , (2) => 2³¹ . 512⁷ . 4³ < 2³¹ . 625⁷ . 5³ ( vì 512⁷ < 625⁷ và 4³ < 5³ )

=> 2¹⁰⁰ < 10³¹

e) Ta có:

2¹⁰⁰ = (2¹⁰)¹⁰ = 1024¹⁰

10³⁰ = (10³)¹⁰ = 1000¹⁰
Vì 1024¹⁰ > 1000¹⁰ => 2¹⁰⁰ > 10³⁰

Ta có :

         = (2³)¹⁰ + (3³)¹⁰ + (4³)¹⁰

         = 8¹⁰ + 27¹⁰ + 64¹⁰

         = ( 3²)¹⁰ + (6²)¹⁰ + (8²)¹⁰

         ⁼ 9¹⁰ + 36¹⁰ + 64¹⁰

Ta thấy: 8¹⁰ + 27¹⁰ + 64¹⁰ < 9¹⁰ + 36¹⁰ + 64¹⁰

\(\Rightarrow\) 2³⁰ + 3³⁰ + 4³⁰ < 3²⁰ + 6²⁰ + 8²⁰

a)10²⁰ và 90¹⁰

Ta có: 10²⁰ =(10²)¹⁰=100¹⁰

Vì 100¹⁰>90¹⁰ nên 10²⁰>90¹⁰

b)(-5)³⁰ và (-3)³⁰

Vì -5<-3 nên (-5)³⁰<(-3)³⁰

b) (-5)³⁰ và (-3)³⁰

Ta có: (-5)³⁰=5³⁰

(-3)³⁰=3³⁰

Vì 5>3 nên (-5)³⁰>(-3)³⁰

a)Ta có : \(10^{20}=\left(10^2\right)^{10}=100^{10}\)

Vì \(100^{10}>90^{10}\Rightarrow10^{20}>90^{10}\)

b)Ta có: \(\left(-5\right)^{30}=\left[\left(-5\right)^2\right]^{15}=25^{15}\)

\(\left(-3\right)^{30}=\left[\left(-3\right)^2\right]^{15}=9^{15}\)

Vì \(25^{15}>9^{15}\Rightarrow\left(-5\right)^{30}>\left(-3\right)^{30}\)

c)Ta có :\(32^9=\left(2^5\right)^9=2^{45}>2^{42}=\left(2^6\right)^7=64^7\\ 6^{13}< 6^{14}=\left(6^2\right)^7=36^7\)

Vì \(2^{45}>64^7>36^7>6^{13}\Rightarrow2^{45}>6^{13}\Rightarrow32^9>6^{13}\)

Vậy \(\left(-32\right)^9>\left(-6\right)^{13}\)

ta có \(2^{30}=\left(2^3\right)^{10}=8^{10}\)

        \(3^{30}=\left(3^3\right)^{10}=27^{10}\)

        \(4^{30}=\left(4^3\right)^{10}=64^{10}\)

   ta có       \(3^{20}=\left(3^2\right)^{10}=9^{10}\)

                  \(6^{20}=\left(6^2\right)^{10}=36^{10}\)

                   \(8^{20}=\left(8^2\right)^{10}=64^{10}\)

              \(\Rightarrow2^{30}+3^{30}+4^{30}=8^{10}+27^{10}+64^{10}\)

            \(\Rightarrow3^{20}+6^{20}+8^{20}=9^{10}+36^{10}+64^{10}\)

       Xét        \(8^{10}

So sánh 2^20+3^30+4^30 và3.24^10