9 . 9 . 9 . 9 = 9⁴

tk nhé

9 . 9 . 9 . 9 = \(9^4\)

~ ai tk mk mk tk lại cho nha ~

9.9.9.9=\(9^4\)

\(a.a.a.a.a.a.a.a.a=a^9\)

k mk nhé

a.a.a.a.a.a.a.a.a = a⁹

k nhé

lũy thừa dưới dạng \(a^9\)

x¹.x⁵.x⁹.x¹³.......x²⁰¹⁷ 

= x^{1+5+9+13+...+2017}

= x⁵⁰⁹⁵⁴⁵

Theo bài ra ta có : x¹.x⁵.x⁹.....x²⁰¹⁷ => x¹.x⁵.x⁹.....x²⁰¹⁷ =x^{1+5+9+.....+2017}

Đặt 1 + 5 + 9 + ..... + 2017 là A

A = 1 + 5 + 9 + ..... + 2017 

A có số số hạng là :

     ( 2017 - 1) : 4 + 1 = 505 số số hạng

Tổng A là :

  ( 2017 + 1 ) nhân 505 : 2 = 509545

 Vậy kết quả là: x⁵⁰⁹⁵⁴⁵

2^5.2^4:2^2.1/16=(2^7)/16

\(\frac{32\cdot2^4}{2^2\cdot\frac{1}{16}}\)

\(=\frac{2^5\cdot2^4}{\frac{4}{16}}\)

\(=\frac{2^9}{\frac{1}{4}}\)

\(=2^9\cdot4\)

\(=2^9\cdot2^2\)

\(=2^{11}\)

=32 X 16 : ( 4 X 1/16)=2048

    32 . 2⁴ : ( 2² . 1/16 ) 

=  2⁵ . 2⁴ : 1/2²

=  2⁹ . 2²

= 2¹¹

\(\frac{2^7}{16}\)

\(2^{11}\)

\(a,3^2\cdot\frac{1}{243}=3^2\cdot\frac{1}{3^5}=\frac{1}{3^3}=\frac{1^3}{3^3}=\left(\frac{1}{3}\right)^3\)

\(b,81^2\cdot\frac{1}{3^3}=\left(3^4\right)^2\cdot\frac{1}{3^3}=3^8\cdot\frac{1}{3^3}=3^5\)

a) \(3^2.\frac{1}{243}=\frac{1.3^3}{243}=\frac{3^2}{243}=\frac{3^2}{3^5}=\frac{1}{3^3}=\frac{1}{27}\)

b) \(81^2.\frac{1}{3^3}=\frac{1.81^2}{3^3}=\frac{81^2}{3^3}=\frac{3^8}{3^3}=3^5=243\)

\(a,5\cdot5\cdot5\cdot5=5^4\\b,9\cdot9\cdot9\cdot9\cdot9\cdot9\cdot9=9^7\\c,7\cdot7\cdot7\cdot7\cdot7=7^5\\d,a\cdot a\cdot a\cdot a\cdot a\cdot a\cdot a\cdot a=a^8\)

a/ 5⁴

b/9⁷

c/7⁵

d/a⁸

\(a\))\(5.5.5.5=5^4\) 

\(b\))\(9.9.9.9.9.9.9=9^7\) 

\(c\))\(7.7.7.7.7=7^5\)

\(d\))\(a.a.a.a.a.a.a.a=a^8\)