\(\dfrac{11\cdot9^{11}\cdot3^7-27^{10}}{4\cdot81^1}\)

\(=\dfrac{11\cdot3^{22}\cdot3^7-\left(3^3\right)^{10}}{4\cdot3^4}\)

\(=\dfrac{11\cdot3^{29}-3^{30}}{4\cdot3^4}\)

\(=\dfrac{3^{29}\cdot\left(11-3\right)}{4\cdot3^4}\)

\(=\dfrac{3^{29}\cdot8}{4\cdot3^4}\)

\(=3^{25}\cdot2\)

@ Huỳnh Thanh Phong

Sao lại lấy 3²⁹ là thừa số chung ạ

x=1;

x=1;

x=1.

k cho mình nhé.

dễ

\(\frac{11.9^{11}.3^7-9^{15}}{\left(2.3^{14}\right)^2}\)

= \(\frac{11.\left(3^2\right)^{11}.3^7-\left(3^2\right)^{15}}{2^2.3^{28}}\)

= \(\frac{11.3^{22}.3^7-3^{30}}{4.3^{28}}\)

= \(\frac{11.3^{29}-3^{29}.3}{4.3^{28}}\)

= \(\frac{\left(11-3\right).3^{29}}{4.3^{28}}\)

= \(\frac{8.3^{29}}{4.3^{28}}\)

= 2 . 3

= 6

A = 51 − 314 + 213 ; A = 51 − 314 − 213 B = 51 + 213 − 314 ; C = 51 + 314 − 213 ; D = 51 + 213 − 314 ; D = 51 + 213 − 314 E = 51 − 314 − 213 ; E = 51 − 314 + 213 F = 51 − 213 − 314. V ậ y   A = F ;   B = E = D

câu hỏi tương tự có nhé

5.7.11.13.37-10101/1212120+40404

=3.5.7.11.13.37-10101/1212120.1/10+40404 (vì 1/1212120=1/121212.1/10)

= 3.37.7.11.13.5-101010/121212.1/100+40404

=111.1001.5-5/6.1/100+40404

=151515.5-250/3

=595959-250/3

=1787876/3

Ta có:

 S = 1.1!+2.2!+3.3!+.....+100.100!

 S = (1+1-1).1!+(2+1-1).2!+...+(100+1-1).100!

 S = 2!-1+3!-2!+4!-3!+...+101!-100!

 S = 101!+(100!-100!)+(99!-99!)+...+(2!-2!)-1

 S = 101!-1

k cho mk nha bạn 

                       S=101! - 1