\(\left(\dfrac{1}{2}\right)^3.\left(\dfrac{3}{2}\right)^3\)

=\(\dfrac{1}{8}.\dfrac{27}{8}\)

=\(\dfrac{27}{64}\)

27 < 81³ : 3^x <243

⇒ 3³ < (3⁴)³ : 3^x <3⁵

⇒ 3³ < 3¹² : 3^x <3⁵

⇒ 3³ < 3^12-x <3⁵

⇒ 3 <12-x < 5

⇒ -9 < x < 7

1) ab=2 (I); bc=3 (II); ca=54 (III)

Lấy (I).(II).(III) ⇒ a² . b² . c² = 324 ⇒ abc = ±18

(II) ⇒ a= ±6 ; (I) ⇒ b= ±1/3 ; (II) ⇒ c= ±9

2) ab=5/3 (I); bc=4/5 (II); ca=3/4 (III)

Lấy (I).(II).(III) ⇒ a² . b² . c² = 1 ⇒ abc = ±1

(II) ⇒ a= ±5/4 ; (I) ⇒ b= ±4/3 ; (II) ⇒ c= ±3/5

3) a(a+b+c)= -12 (I)

    b(a+b+c)= 18 (II)

    c(a+b+c)= 30 (III)

Lấy (I)+(II)+(III) ⇒ (a+b+c)² = 36 ⇒ a+b+c = ±6

TH1 : a=6 ⇒ a= -12/6 = -2 ; b= 18/6 = 3 ; c= 30/6 = 5

TH2 : a=-6 ⇒ a= -12/-6 = 2 ; b= 18/-6 = -3 ; c= 30/-6 = -5

\(A=3+3^2+3^3+...+3^{2015}\)

\(\Rightarrow3A=3^2+3^3+...+3^{2015}+3^{2016}\)

\(\Rightarrow3A-A=\left(3^2+3^3+...+3^{2016}\right)-\left(3+3^2+3^3+...+3^{2015}\right)\)

\(\Rightarrow2A=\left(3^2-3^2\right)+\left(3^3-3^3\right)+...+\left(3^{2016}-3\right)\)

\(\Rightarrow2A=3^{2016}-3\)

\(\Rightarrow A=\dfrac{3^{2016}-3}{2}\)

Ta có: \(2A+3=3^n\)

\(\Rightarrow2\cdot\dfrac{3^{2016}-3}{2}+3=3^n\)

\(\Rightarrow3^{2016}-3+3=3^n\)

\(\Rightarrow3^{2016}=3^n\)

\(\Rightarrow n=2016\)

81^7 - 27^9 - 9^13
= (3^4)^7 - (3^3)^9 - (3^2)^13
= 3^28 - 3^27 - 3^26
= (3^26.3^2) - (3^26.3^1) - (3^26.1)
= 3^26.(9 - 3 - 1)
= 3^22.(3^4.5)
= 3^22.405 chia hết cho 405
=> 81^7 - 27^9-9^13 chia hết cho 405

Không chia hết đâu bạn ơi

Bạn xem lại đề. Thay $n=1$ thì biểu thức không chia hết cho 7 nhé.

\(\left(\dfrac{2}{5}-\dfrac{1}{3}\right)^2\)

=  \(\left(\dfrac{1}{15}\right)^2\)

=  \(\dfrac{1}{225}\)

\(=3^2.3^n-2^2.2^n+3^n-2^n=10.3^n-5.2^n=\)

\(=10.3^n-10.2^{n-1}=10.\left(3^n-2^n\right)⋮10\)

\(\left(3b^2\right)^2-b^3.\left(1-5b\right)\)

\(=9b^4-b^3+5b^4\)

\(=14b^4-b^3\)