5^{x - 1} = 5²

⇒ x - 1 = 2

⇒ x = 2 + 1

⇒ x = 3

Vậy x = 3

1.371742e+190 đấy :)

sosososos dịch ra là sos os os os

hehe

\(1^{2014}=1\) (1)

\(2^{2014}=2^{4.503+2}=\left(2^4\right)^{503}.2^2=16^{503}.4=\overline{...6}.4=\overline{...4}\) (2)

\(3^{2014}=3^{4.503+2}=\left(3^4\right)^{503}.3^2=81^{503}.9=\overline{...1}.9=\overline{...9}\) (3)

\(4^{2014}=4^{2.1007}=\left(4^2\right)^{1007}=16^{1007}=\overline{...6}\) (4)

\(5^{2014}=\overline{...5}\) (5)

\(6^{2014}=\overline{...6}\) (6)

\(7^{2014}=7^{4.503+2}=\left(7^4\right)^{503}.7^2=2401^{503}.49=\overline{...1}.49=\overline{...9}\) (7)

\(8^{2014}=8^{4.503+2}=\left(8^4\right)^{503}.8^2=4096^{503}.64=\overline{...6}.64=\overline{...4}\) (8)

Lấy `(1) + (2)` được số tận cùng là 5 chia hết cho 5

Lấy `(3) + (4)` được số tận cùng là 5 chia hết cho 5

Lấy `(6) + (8)` được số tận cùng là 0 chia hết cho 5

Mà `(5)` có tận cùng là chữ số 5 chia hết cho 5 

Nên  `(1) + (2) +  (3) + (4) + (5) + (6) + (8)` cho kết quả là số chia hết cho 5

Ta có: `(7)` tận cùng là chữ số 9 mà 9 chia 5 dư 4

Vậy phép tính trên chia 5 dư 4

a)
\(8.8.64.2.2\\ =\left(8.2\right).\left(8.2\right).64\\ =16.16.64\\ =4^2.4^2.4^3\\ =4^{2+2+3}\\ =4^7\)

b)
\(27.9.9.3.3\\ =27.\left(9.3\right).\left(9.3\right)\\ =27.27.27\\ =27^{1+1+1}\\ =27^3\)

a) 8.8.64.2.2

= 2³.2³.2⁵.2²

= 2³⁺³⁺⁵⁺²

= 2¹³

b) 27.9.9.3.3

= 3³.3².3².3²

= 3³⁺²⁺²⁺²

= 3⁹

\(a,27\cdot3^{n-1}=81\\ \Rightarrow3^{n-1}=81:27\\ \Rightarrow3^{n-1}=3\\ \Rightarrow n-1=1\\ \Rightarrow n=2\)

\(b,5\cdot5^{6-n}=625\\ \Rightarrow5^{6-n}=625:5\\ \Rightarrow5^{6-n}=125\\ \Rightarrow5^{6-n}=5^3\\ \Rightarrow6-n=3\\ \Rightarrow n=3\)

\(a,27\cdot3^{n-1}=81\\ =>3^3\cdot3^{n-1}=81\\ =>3^{n-1+3}=3^4\\ =>3^{n+2}=3^4\\ =>n+2=4\\ =>n=4-2\\ =>n=2\\ b,5\cdot5^{6-n}=625\\ =>5^{1+6-n}=5^4\\ =>5^{7-n}=5^4\\ =>7-n=4\\ =>n=7-4\\=>n=3\)

\(\left(1-\dfrac{1}{4}\right)\left(1-\dfrac{1}{5}\right)\left(1-\dfrac{1}{6}\right)\left(1-\dfrac{1}{7}\right)\left(1-\dfrac{1}{8}\right)\left(1-\dfrac{1}{9}\right)=\dfrac{a}{75}\\ =>\dfrac{3}{4}\cdot\dfrac{4}{5}\cdot\dfrac{5}{6}\cdot\dfrac{6}{7}\cdot\dfrac{7}{8}\cdot\dfrac{8}{9}=\dfrac{a}{75}\\ =>\dfrac{3\cdot\left(4\cdot5\cdot6\cdot7\cdot8\right)}{\left(4\cdot5\cdot6\cdot7\cdot8\right)\cdot9}=\dfrac{a}{75}\\ =>\dfrac{3}{9}=\dfrac{a}{75}\\ =>\dfrac{a}{75}=\dfrac{1}{3}\\ =>a=\dfrac{1}{3}\cdot75\\ =>a=25\)

Vậy: ...

\(a,7\cdot7\cdot7\cdot7\\ =7^1\cdot7^1\cdot7^1\cdot7^1\\ =7^{1+1+1+1}=7^4\\ b,7\cdot35\cdot7\cdot25\\ =7\cdot\left(5\cdot7\right)\cdot7\cdot\left(5^2\right)\\ =7\cdot5\cdot7\cdot7\cdot5^2\\ =\left(7\cdot7\cdot7\right)\cdot\left(5\cdot5^2\right)\\ =\left(7^1\cdot7^1\cdot7^1\right)\cdot\left(5^1\cdot5^2\right)\\ =7^{1+1+1}\cdot5^{1+2}\\ =7^3\cdot5^3\\ c,2\cdot3\cdot8\cdot12\cdot24\\ =2\cdot3\cdot2^3\cdot\left(2^2\cdot3\right)\cdot\left(2^3\cdot3\right)\\ =\left(2\cdot2^3\cdot2^2\cdot2^3\right)\cdot\left(3\cdot3\cdot3\right)\\ =\left(2^1\cdot2^3\cdot2^2\cdot2^3\right)\cdot\left(3^1\cdot3^1\cdot3^1\right)\\ =2^{1+3+2+3}\cdot3^{1+1+1}\\ =2^9\cdot3^3\\ d,x\cdot x\cdot y\cdot y\cdot x\cdot y\cdot x\\ =\left(x\cdot x\cdot x\cdot x\right)\cdot\left(y\cdot y\cdot y\right)\\ =\left(x^1\cdot x^1\cdot x^1\cdot x^1\right)\cdot\left(y^1\cdot y^1\cdot y^1\right)\\ =x^{1+1+1+1}\cdot y^{1+1+1}\\ =x^4\cdot y^3\)

a) Ta có:
\(x-y⋮7\)
Vì \(21x⋮7\) nên:
\(x-y+21x⋮\\ \Rightarrow22x-y⋮7\)
Vậy...

b) Ta có:
\(x-y⋮7\)
Vì \(7x⋮7\) và \(21y⋮7\) nên:
\(x-y+7x+21y⋮\\ \Rightarrow8x+20y⋮7\)
Vậy...

c) Ta có:
\(x-y⋮7\\ \Rightarrow11.\left(x-y\right)⋮7\\ \Rightarrow11x-11y⋮7\)
Vì \(21y⋮7\) nên:
\(11x-11y+21y⋮\\ \Rightarrow11x+10y⋮7\)
Vậy...

Ta có `x - y ⋮ 7`

`=>x-y=7k(k∈N)` 

`=>x=7k+y`

a) `22x-y`

`=22(7k+y)-y`

`=7k*22+22y-y`

`=7k*22+21y`

`=7*(22k+3y)⋮7` 

b) `8x+20y`

`=8(7k+y)+20y`

`=56k+8y+20y`

`=56k+28y`

`=7*(8k+4y)⋮7`

c) `11x+10y`

`=11(7k+y)+10y`

`=77k+11y+10y`

`=77k+21y`

`=7*(11k+3y)⋮7` 

\(a,6^8\cdot6^8\cdot6^5\\ =6^{8+8+5}\\ =6^{21}\\ b,25^2\cdot5^3\cdot5\\ =\left(5^2\right)^2\cdot5^3\cdot5\\ =5^{2\cdot2}\cdot5^3\cdot5\\ =5^4\cdot5^3\cdot5\\ =5^{4+3+1}\\ =5^8\)

\(a,9^{11}:9^2:9^3\\ =9^{11-2-3}\\ =9^6\\ =\left(3^2\right)^6\\ =3^{2\cdot6}\\ =3^{12}\\ b,3^{21}:3^7:3\\ =3^{21-7-1}\\ =3^{13}\)

a) (Sửa đề) 
\(9^{11}:9^2:9:3\\ =\left(3^2\right)^{11}:\left(3^2\right)^2:3^2:3\\ =3^{22}:3^4:3^2:3\\ =3^{22}:\left(3^4.3^2.3\right)\\ =3^{22}:3^7\\ =3^{15}\)

b) 
\(3^{21}:3^7:3\\ =3^{21}:\left(3^7.3\right)\\ =3^{21}:3^8\\ =3^{13}\)