a) Đặt A = \(6^5.5-3^5\)

\(=\left(2.3\right)^5.5-3^5\)

\(=2^5.3^5.5-3^5\)

\(=3^5.\left(2^5.5-1\right)\)

\(=3^5.\left(32.5-1\right)\)

\(=3^5.159\)

\(=3^5.3.53⋮53\)

Vậy \(A⋮53\)

b) Đặt \(B=2+2^2+2^3+...+2^{120}\)

\(=\left(2+2^2\right)+\left(2^3+2^4\right)+...+\left(2^{119}+2^{120}\right)\)

\(=2.\left(1+2\right)+2^3.\left(1+2\right)+...+2^{119}.\left(1+2\right)\)

\(=2.3+2^3.3+...+2^{119}.3\)

\(=3.\left(2+2^3+...+2^{59}\right)⋮3\)

Vậy \(B⋮3\)

\(B=\left(2+2^2+2^3\right)+\left(2^4+2^5+2^6\right)+...+\left(2^{118}+2^{119}+2^{120}\right)\)

\(=2.\left(1+2+2^2\right)+3^4.\left(1+2+2^2\right)+...+2^{118}.\left(1+2+2^2\right)\)

\(=2.7+2^4.7+...+2^{118}.7\)

\(=7.\left(2+2^4+...+2^{118}\right)⋮7\)

Vậy \(B⋮7\)

\(B=\left(2+2^2+2^3+2^4+2^5\right)+\left(2^6+2^7+2^8+2^9+2^{10}\right)\)

\(+...+\left(2^{116}+2^{117}+2^{118}+2^{119}+2^{120}\right)\)

\(=2.\left(1+2+2^2+2^3+2^4\right)+2^6.\left(1+2+2^2+2^3+2^4\right)\)

\(+2^{116}.\left(1+2+2^2+2^3+2^4\right)\)

\(=2.31+2^6.31+...+2^{116}.31\)

\(=31.\left(2+2^6+...+2^{116}\right)⋮31\)

Vậy \(B⋮31\)

\(B=\left(2+2^2+2^3+2^4+2^5+2^6+2^7+2^8\right)+\left(2^9+2^{10}+2^{11}+2^{12}+2^{13}+2^{14}+2^{15}+2^{16}\right)\)

\(+...+\left(2^{113}+2^{114}+2^{115}+2^{116}+2^{117}+2^{118}+2^{119}+2^{120}\right)\)

\(=2.\left(1+2+2^2+2^3+2^4+2^5+2^6+2^7\right)+2^9.\left(1+2+2^2+2^3+2^4+2^5+2^6+2^7\right)\)

\(+...+2^{113}.\left(1+2+2^2+2^3+2^4+2^5+2^6+2^7\right)\)

\(=2.255+2^9.255+...+2^{113}.255\)

\(=255.\left(2+2^9+...+2^{113}\right)\)

\(=17.15.\left(2+2^9+...+2^{113}\right)⋮17\)

Vậy \(B⋮17\)

c) Đặt C = \(3^{4n+1}+2^{4n+1}\)

Ta có:

\(3^{4n+1}=\left(3^4\right)^n.3\)

\(2^{4n}=\left(2^4\right)^n.2\)

\(3^4\equiv1\left(mod10\right)\)

\(\Rightarrow\left(3^4\right)^n\equiv1^n\left(mod10\right)\equiv1\left(mod10\right)\)

\(\Rightarrow3^{4n+1}\equiv\left(3^4\right)^n.3\left(mod10\right)\equiv1.3\left(mod10\right)\equiv3\left(mod10\right)\)

\(\Rightarrow\) Chữ số tận cùng của \(3^{4n+1}\) là \(3\)

\(2^4\equiv6\left(mod10\right)\)

\(\Rightarrow\left(2^4\right)^n\equiv6^n\left(mod10\right)\equiv6\left(mod10\right)\)

\(\Rightarrow2^{4n+1}\equiv\left(2^4\right)^n.2\left(mod10\right)\equiv6.2\left(mod10\right)\equiv2\left(mod10\right)\)

\(\Rightarrow\) Chữ số tận cùng của \(2^{4n+1}\) là \(2\)

\(\Rightarrow\) Chữ số tận cùng của C là 5

\(\Rightarrow C⋮5\)

d) Đặt \(D=75+\left(4^{2006}+4^{2005}+4^{2004}+...+1\right).25\)

Đặt \(E=4^{2006}+4^{2005}+4^{2004}+...+1\)

\(\Rightarrow4E=4^{2007}+4^{2006}+4^{2005}+...+4\)

\(\Rightarrow3E=4E-E\)

\(=\left(4^{2007}+4^{2006}+4^{2005}+...+4\right)-\left(4^{2006}+4^{2005}+4^{2004}+...+1\right)\)

\(=4^{2007}-1\)

\(\Rightarrow E=\dfrac{\left(4^{2007}-1\right)}{3}\)

\(\Rightarrow D=75+\dfrac{4^{2007}-1}{3}.25\)

Ta có:

\(4^{2007}=\left(4^2\right)^{1003}.4\)

\(4^2\equiv6\left(mod10\right)\)

\(\left(4^2\right)^{1003}\equiv6^{1003}\left(mod10\right)\equiv6\left(mod10\right)\)

\(\Rightarrow4^{2007}\equiv\left(4^2\right)^{1003}.4\left(mod10\right)\equiv6.4\left(mod10\right)\equiv4\left(mod10\right)\)

\(\Rightarrow\) Chữ số tận cùng của \(4^{2007}\) là 4

x=5 nha bạn

\(x\in\){ 5 }

Tk mk mk tk lại ! ^^ 

a, đề k rõ :v

b, (x + 2)⁴ = 625

=> (x + 2) = (+5)⁴  

=> x + 2 = + 5

=> x = 3 hoặc x = -7

vậy_

a, \(4^{n+2}\) = 64

\(4^{n+2}\) = \(4^3\)

n + 2 = 3

n = 3 -2

n = 1

a, \(\left(x+2\right)^4\) = 625

\(\left(x+2\right)^4\) = \(5^4\)

x +2 = 5

x= 5 - 2

x= 3

Tk mk nha

(x+2)⁴=625

(x+2)⁴=5⁴

x+2=5(vì  số mũ 4>0)

x=3

kb nha

\(A=\left\{x\inℕ|x\le6\right\}=\left\{0;1;2;3;4;5;6\right\}=\left\{0;2;5;4;m;6;n\right\}\)

Vậy m và n phải mang giá trị của 1 hoặc 3

\(\Rightarrow\orbr{\begin{cases}m=1,n=3\\m=3,n=1\end{cases}}\)

\(A=\left\{x\in N|x\le6\right\}\)

\(\Rightarrow A=\left\{0;1;2;3;4;5;6\right\}\)

\(A=\left\{0;2;5;4;m;6;n\right\}\)

\(A=\text{{}0;m/n;2;n/m;4;5;6\)

\(m/n-1=0\);\(n/m-1=2\)

Vì m hoặc n có thể đổi chỗ cho nhau,nên thỏa mãn điều kiện:

\(\hept{\begin{cases}m=1;m=3\\n=1;n=3\end{cases}}\)

a)X × 7 /7=7

\(\Rightarrow x\cdot7=7\cdot7\)

\(\Rightarrow x\cdot7=49\)

\(\Rightarrow x=7\)

b)X / X × 6=1/6

\(\Rightarrow6\cdot x=1\cdot6\cdot x\)

\(\Rightarrow6x=6x\)

Vậy ta có mọi x thuộc Z thỏa mãn 

M.n giúp mk vs ạ.gấp ạ

Mk cảm ơn nhiều

(2x - 6)^2 = (2x - 6)^4

=> (2x - 6)^4 - (2x - 6)^2 = 0

=> (2x - 6)^2.[(2x - 6)^2 - 1] = 0

=> (2x - 6)^2 = 0 hoặc (2x - 6)^2 - 1 = 0

=> 2x - 6 = 0 hoặc (2x - 6)^2 = 1

=> 2x = 6 hoặc 2x - 6 = + 1 

=> x = 3  hoặc 2x = 7 hoặc 2x = 5

=> x = 3 hoặc x = 7/2 hoặc x = 5/2

mà x thuộc N

=> x = 3

x=2,5

hk tốt!

thanks