Ta có: 27^15 = (3^3)^15 = 3^45

81^11 = (3^4)^11 = 3^44

Vì 45 > 44 nên 3^45 > 3^44

Vậy 27^15 > 81^11

Ta có: \(27^{15}=\left(3^3\right)^{15}=3^{45}\)

           \(81^{11}=\left(3^4\right)^{11}=3^{44}\)

Vì \(45>44\) nên \(27^{15}>81^{11}\)

Ta có : 27 ^15 = (3^3)^15=3^45 ; 81 ^11 = (3^4)^11 = 3^44

Vì 3^45 > 3^44 ( Vì 2 lũy thừa có cùng cơ số > 1 ; 45>44) nên suy ra 27^5 > 81 ^11

27/11>81/8 nha bạn!

\(27^{11}=\left(3^3\right)^{11}=3^{33}\)

\(81^8=\left(3^4\right)^8=3^{32}\)

\(3^{33}>3^{32}\Rightarrow27^{11}>81^8\)

27¹¹ = (3³)¹¹ = 3³³

81⁸ = (3⁴)⁸ = 3³²

Mà: 33 > 32

=> 3³³ > 3³²

=> 27¹¹ > 81⁸ 

a/

\(27^{81}=\left(3^3\right)^{81}=3^{241}\)

\(81^{27}=\left(3^4\right)^{27}=3^{108}\)

\(\Rightarrow27^{81}=3^{241}>3^{108}=81^{27}\)

b/

\(5^{60}=\left(5^3\right)^{20}=125^{20}\)

\(7^{40}=\left(7^2\right)^{20}=49^{20}\)

\(\Rightarrow5^{60}=125^{20}>49^{20}=7^{40}\)

c/

\(11^{102}=\left(11^2\right)^{51}=121^{51}>121^{50}>99^{50}\)

81/8 lớn hơn vì vừa nhìn vào đã thấy , 81 : 8 =10 ( dư ) 1 . 27 : 11 = 2 ( dư 5 ) .

27/11 và 81/8

MSC:88

27/11=216/88

81/8=891/88

Vì: 216/88 < 891/88

Nên: 27/11 < 81/8

\(a,81^3=\left(9^2\right)^3=9^6\)

Vì \(9^{27}>9^6\) nên \(9^{27}>81^3\)

\(b,5^{14}=\left(5^2\right)^7=25^7\)

Vì \(25^7< 27^7\) nên \(5^{14}< 27^7\)

\(c,10^{30}=\left(10^3\right)^{10}=1000^{10}\)

\(2^{100}=\left(2^{10}\right)^{10}=1024^{10}\)

Vì \(1000^{10}< 1024^{10}\) nên \(10^{30}< 2^{100}\)

Ta có :

27³⁰ = (3³)³⁰ = 3 ^{3 . 30} = 3⁹⁰ (1)

81²² = (3⁴)²² = 3^{4 . 22} = 3⁸⁸ (2)

Từ (1) và (2) => 3⁹⁰ > 3⁸⁸ <=> 27³⁰ > 81²²

~Study well~

#Shizu

Trả lời

27³⁰ = (3³)³⁰ = 3⁹⁰

81²² = (3⁴)²² = 3⁸⁸

Học tốt !

Ta có :

27³⁰ = ( 3³ )³⁰ = 3⁹⁰

81²² = ( 3⁴ )²² = 3⁸⁸

Vì 90 > 88 nên 3⁹⁰ > 3⁸⁸ hay 27³⁰ > 81²²

\(\left(\dfrac{1}{81}\right)\) mũ mấy em

Ta có:

\(\left(\dfrac{1}{27}\right)^{10}=\left(\dfrac{1}{3^3}\right)^{10}=\left(\dfrac{1}{3}\right)^{30}\)

\(\left(\dfrac{1}{81}\right)^7=\left(\dfrac{1}{3^5}\right)^7=\left(\dfrac{1}{3}\right)^{35}\)

Vì \(\left(\dfrac{1}{3}\right)^{30}< \left(\dfrac{1}{3}\right)^{35}\)

⇒\(\left(\dfrac{1}{27}\right)^{10}< \left(\dfrac{1}{81}\right)^7\)

\(\left(\dfrac{1}{27}\right)^{10}=\dfrac{1}{27^{10}}=\dfrac{1}{\left(3^3\right)^{10}}=\dfrac{1}{3^{30}}\)

\(\left(\dfrac{1}{81}\right)^7=\dfrac{1}{81^7}=\dfrac{1}{\left(3^4\right)^7}=\dfrac{1}{3^{28}}\)

Do \(3^{30}>3^{28}\Leftrightarrow\dfrac{1}{3^{30}}< \dfrac{1}{3^{28}}\)

\(\Leftrightarrow\left(\dfrac{1}{27}\right)^{10}< \left(\dfrac{1}{81}\right)^7\)

Ta có:

\(\left(\dfrac{1}{27}\right)^{10}=\left(\dfrac{1}{3^3}\right)^{10}=\left(\dfrac{1}{3}\right)^{30}\)

\(\left(\dfrac{1}{81}\right)^7=\left(\dfrac{1}{3^5}\right)^7=\left(\dfrac{1}{3}\right)^{35}\)

Vì \(\left(\dfrac{1}{3}\right)^{35}>\left(\dfrac{1}{3}\right)^{30}\)

⇒\(\left(\dfrac{1}{27}\right)^{10}< \left(\dfrac{1}{81}\right)^7\)