\(\left(\frac{27}{64}\right)^{15}=\frac{\left(3^3\right)^{15}}{\left(2^6\right)^{15}}=\frac{3^{45}}{2^{90}}=\left(\frac{3}{2^2}\right)^{45}\)

\(\left(\frac{81}{256}\right)^{10}=\frac{\left(3^4\right)^{10}}{\left(2^8\right)^{10}}=\frac{3^{40}}{2^{80}}=\left(\frac{3}{2^2}\right)^{40}\)

Do \(\left(\frac{3}{2^2}\right)^{45}

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

\(a,64^5=\left[8^2\right]^5=8^{10}\)

Giữ nguyên \(11^{10}\)

Mà \(8< 11\)=> \(8^{10}< 11^{10}\)hay \(64^5< 11^{10}\)

\(b,81^7=\left[9^2\right]^7=9^{14}\)

Giữ nguyên \(7^{14}\)

Mà \(9>7\)=> \(9^{14}>7^{14}\)hay \(81^7>7^{14}\)

c, Vì \(244>80\)=> \(244^{11}>80^{11}\)

d, Tương tự

a) 64⁵ và 11¹⁰

Ta có : 64⁵ = (8²)⁵ = 8^2.5 = 8¹⁰

Vì 8¹⁰ < 11¹⁰ nên 64⁵ < 11¹⁰

b) 81⁷ và 7¹⁴

Ta có : 81⁷ = (9²)⁷ = 9^2.7 = 9¹⁴

Vì 9¹⁴ > 7¹⁴ nên 81⁷ > 7¹⁴

c) 244¹¹ và 80¹¹

Vì 244 > 80 và số mũ bằng nhau nên 244¹¹ > 80¹¹

=))

d) Ta có:

Vì 62<64 =>  \(62^{15}< 64^{15}\)

34>32 => \(34^{18}>32^{18}\)

Mà \(32^{18}=2^{5.18}=2^{90}=2^{6.15}=64^{15}\)

=> \(34^{18}>62^{15}\)

\(\left(\dfrac{1}{81}\right)\) 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 \(\left(\dfrac{1}{3}\right)^{35}>\left(\dfrac{1}{3}\right)^{30}\)

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

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

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 \(10^{30}< 2^{100}\)

a)243⁷=(3⁵)⁷=3³⁵ ; 9¹⁰.27⁵=3³⁰.3¹⁵=3⁴⁵.

Vì 35 < 45 => 3³⁵<3⁴⁵=>243⁷<9¹⁰.27⁵.

b) 15¹¹=3¹¹.5¹¹;81³.125⁵=3¹².5¹⁵.

Vì 3¹¹<3¹² và 5¹¹<5¹⁵ => 3¹¹.5¹¹<3¹².5¹⁵ => 15¹¹ < 81³.125^5​

các bạn làm sao mà viết mũ được vậy

27^15=(3^3)^15=3^45

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

vì 3^45>3^44

=>27^15>81^11

ta có : 27¹⁵= (3³)¹⁵= 3⁴⁵

             81¹¹= (3⁴)¹¹ = 3⁴⁴

mà 45 > 44 => 3⁴⁵ > 3⁴⁴

=>  25¹⁵ > 81¹¹

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

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

\(\Rightarrow3^{45}>3^{44}\)

\(\Rightarrow27^{15}>81^{11}\)