\(25^{15}=\left(5^2\right)^{15}=5^{30}\)

\(8^{10}.3^{30}=\left(2^3\right)^{10}.3^{30}=2^{30}.3^{30}=\left(2.3\right)^{30}=6^{30}\)

Vì 5 < 6 => 5³⁰ < 6³⁰

Vậy \(25^{15}<8^{10}.3^{30}\).

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

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

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

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) 16^30=(2^4)^30=2^120

32^15=(2^5)^15=2^75

Vậy, 16^30>32^15

b) 27^48=(3^3)^48=3^144

243^32=((3^5)^32=3^160

Vậy, 27^48<243^32.

Mình làm được 2 câu trên. Nhớ k cho mình.

\(\frac{13}{15}< \frac{11}{13}\)

\(\frac{8}{5}< \frac{7}{4}\)

\(\frac{8}{9}< \frac{4}{5}\)

\(\frac{7}{15}< \frac{21}{4}\)

Theo thứ tự≥≥≥= k mik nhé!

cái này thì mình ko bít nhưng mình nghĩ chắc bn thích gangaa đúng ko?

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