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\(\left(\frac{1}{16}\right)^{10}\) và \(\left(\frac{1}{2}\right)^{50}\)

Ta có: \(\left(\frac{1}{2}\right)^{50}=\left[\left(\frac{1}{2}\right)^5\right]^{10}=\left(\frac{1}{32}\right)^{10}\)

Do \(\frac{1}{6}>\frac{1}{32}\Rightarrow\left(\frac{1}{6}\right)^{10}>\left(\frac{1}{32}\right)^{10}\)

Vậy \(\left(\frac{1}{16}\right)^{10}>\left(\frac{1}{2}\right)^{50}\)

a) \(10^{20}\) và \(9^{10}\)

Vì 10 > 9 ; 20 > 10

nên \(10^{20}>9^{10}\)

Vậy \(10^{20}>9^{10}\)

b) \(\left(-5\right)^{30}\) và \(\left(-3\right)^{50}\)

Ta có: \(\left(-5\right)^{30}=5^{30}=\left(5^3\right)^{10}=125^{10}\)

           \(\left(-3\right)^{50}=3^{50}=\left(3^5\right)^{10}=243^{10}\)

Vì 243 > 125 nên \(125^{10}< 243^{10}\)

Vậy \(\left(-5\right)^{30}< \left(-3\right)^{50}\)

c) \(64^8\) và \(16^{12}\)

Ta có: \(64^8=\left(4^3\right)^8=4^{24}\)

          \(16^{12}=\left(4^2\right)^{12}=4^{24}\)

Vậy \(64^8=16^{12}\left(=4^{24}\right)\)

d) \(\left(\frac{1}{6}\right)^{10}\) và \(\left(\frac{1}{2}\right)^{50}\)

Ta có: \(\left(\frac{1}{6}\right)^{10}=\left[\left(\frac{1}{2}\right)^4\right]^{10}=\left(\frac{1}{2}\right)^{40}\)

Vì 40 < 50 nên \(\left(\frac{1}{2}\right)^{40}< \left(\frac{1}{2}\right)^{50}\)

Vậy \(\left(\frac{1}{16}\right)^{10}< \left(\frac{1}{2}\right)^{50}\)

tham khảo:

Ta có :2²⁷= 2^9.3=512³

            : 3¹⁸=3^6.3=729³

2²⁷ = (2³)⁹ = 8⁹

3¹⁸ = ( 3²)⁹ = 9⁹

Vì 9 > 8 nên : 9⁹ > 8⁹

Vậy suy ra: 3¹⁸ > 2²⁷

\(8^{15}=\left(2^3\right)^{15}=2^{3.15}=2^{45}\\ 16^4=\left(2^4\right)^4=2^{4.4}=2^{16}\\ 2^{45}>2^{16}\Rightarrow8^{15}>16^4\)

Cảm ơn bạn nhiều.

Ta có :

• 9999=101.99\(\Rightarrow\)9999¹⁰=(101.99)¹⁰=101¹⁰.99¹⁰
• 99²⁰=99¹⁰⁺¹⁰=99¹⁰.99¹⁰

​Vì 101¹⁰>99¹⁰\(\Leftrightarrow\)101¹⁰.99¹⁰>99¹⁰.99¹⁰\(\Leftrightarrow\)9999¹⁰>99²⁰

Vậy 9999¹⁰>99²⁰

99^20 < 9999^10

2²⁰ = (2⁵)⁴ = 32⁴

3¹² = (3³)⁴ = 27⁴

Do 32 > 27 nên 32⁴ > 27⁴

Vậy 2²⁰ > 3¹²

2²⁰ = (2⁵)⁴ = 32⁴

3¹² = (3³)⁴ = 27⁴

Vì 32 > 27 ⇒ 32⁴ > 27⁴ ⇒ 2²⁰ > 3¹²