ko biết

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25^15>(-8)^10 x 3^30

k cho mình nhé

\(25^{15}=5^{30}\)

\(8^{10}\cdot3^{30}=6^{30}\)

Do đó: \(25^{15}< 8^{10}\cdot3^{30}\)

vào câu hỏi tương tự

\(2^{333}< 3^{222}\)

mình cần cách giải

1,10²⁰và 90¹⁰

ta có:+,10²⁰=(10²)¹⁰=100¹⁰

        +,90¹⁰=90¹⁰

vì 100¹⁰>90¹⁰=>10²⁰>90¹⁰

2,(1/16)¹⁰ và (1/2)⁵⁰

ta có:+, (1/16)¹⁰=(1/16)¹⁰

         +,(1/2)⁵⁰=(1/2⁵)¹⁰=(1/32)¹⁰

vì (1/16)¹⁰>(1/32)¹⁰=>(1/16)¹⁰>(1/2)⁵⁰

k mik nhé

\(a,\)  \(10^{20}=10^{10+10}=10^{10}.10^{10}\)

        \(90^{10}=9^{10}.10^{10}\)

  Vì \(10^{10}.10^{10}>9^{10}.10^{10}\)

    \(\Rightarrow10^{20}>90^{10}\)

Vậy \(10^{20}>90^{10}\)

\(b,\)\(\left(\frac{1}{16}\right)^{10}=\frac{1^{10}}{16^{10}}=\frac{1}{\left(4^2\right)^{10}}=\frac{1}{4^{20}}\)

   \(\left(\frac{1}{2}\right)^{50}=\frac{1^{50}}{2^{50}}=\frac{1}{\left(2^2\right)^{25}}=\frac{1}{4^{25}}\)

Vì \(\frac{1}{4^{20}}>\frac{1}{4^{25}}\)

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

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

                         ~~~~~~~~~~Hok tốt~~~~~~~~~~~

\(2^{50}=\left(2^5\right)^{10}=32^{10}\)

\(5^{20}=\left(5^2\right)^{10}=25^{10}\)

Suy ra: 2⁵⁰ > 5²⁰

b)

\(9^{200}=\left(9^2\right)^{100}=81^{100}\)

Suy ra: 99¹⁰⁰ > 81¹⁰⁰

\(5^{202}=\left(5^2\right)^{101}=25^{101}\)

\(2^{505}=\left(2^5\right)^{101}=32^{101}\)

Suy ra: 5²⁰² < 2⁵⁰⁵

a) Ta có: \(3^{40}=\left(3^4\right)^{10}=81^{10}\)

              \(5^{30}=\left(5^3\right)^{10}=125^{10}\)

Vì 125 > 81 => \(125^{10}>81^{10}\) => \(3^{40}>5^{30}\)

b) Ta có: \(5^{303}>5^4\) vì 303 > 4

         Mà: \(5^4>2^4\)  vì 5 > 2

=> \(5^{303}>2^4\)

Ta có:

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

\(\left(-3\right)^{50}=\left(-3^5\right)^{10}=\left(-81\right)^{10}=81^{10}\)

Vì \(125^{10}>81^{10}\)

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

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

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

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

Vì \(\left(-125\right)^{10}< \left(-243\right)^{10}\) nên \(\left(-5\right)^{30}< \left(-3\right)^{50}\)