Ta có : \(\left(x-2\right)^{2016}\)dương

\(\Rightarrow x-3=0\Rightarrow x=3\)

Thay x ta thử :

 \(\left(3-2\right)^{2016}+\left(3-3\right)=1+0=1\)thỏa đề

Vậy \(x=3\)

Đặt \(A=\left(x-2\right)^{2016}+\left(x-3\right)\)

\(x-2< 2\) vì nếu \(x-2\ge2\)

\(\Rightarrow x-3\ge1\)

\(\left(x-2\right)^{2016}>3\)

\(\Rightarrow A=\left(x-2\right)^{2016}+\left(x-3\right)>3\) ( vô lý )

\(\Rightarrow x-2< 2\)

\(\Rightarrow x< 4\)

Với \(x=0\)

\(\Rightarrow A=\left(x-2\right)^{2016}+\left(x-3\right)=2^{2016}-3>3\)

Với \(x=1\)

\(\Rightarrow A=\left(x-2\right)^{2016}+\left(x-3\right)< 0< 3\)

Với \(x=2\)

\(\Rightarrow A=\left(x-2\right)^{2016}+\left(x-3\right)=0-1< 3\)

Với \(x=3\)

\(\Rightarrow A=\left(x-2\right)^{2016}+\left(x-3\right)=1+0< 3\)

Do đó không có \(x\in N\) thỏa mãn.

Hơi dài đấy. Bạn làm từng tường hợp nhé

giai dum di ma mai nop roi 

\(x^{2016}=x^5\)

=>\(x^{2016}-x^5=0\)

=>\(x^5\left(x^{2011}-1\right)=0\)

=>x⁵=0 hoặc x²⁰¹¹-1=0

=>x=0 hoặc x=1

x=0 x=1

x=0 hoặc x=1

Vì: 1²⁰¹⁶=1

1⁵⁼¹

0²⁰¹⁶=0

0⁵=0

x+(x+1)+(x+2)+(x+3)+.......+(x+30)=1240 

\(\Leftrightarrow\left(x+x+x.+...x\right)+\left(1+2+3...+30\right)=1240\)

\(\Rightarrow30x+465=1240\)

\(\Rightarrow30x=1240-465=775\)

\(\Rightarrow30x=775\)

\(V\text{ậy}x=\frac{155}{6}\)

1+2+3+.....+x=210

\(\left(1+x\right).x=210\)

\(\Rightarrow x=14\)

x+(x+1)+(x+2)+...+(x+30)=1240

=>x+x+1+x+2+...+x+30=1240

=>(x+x+x+...+x)+(1+2+...+30)=1240

=>31x+[(30-1):1+1] . (30+1) :2=1240

=>31x+30.31:2=1240

=>31x+15.31=1240

=>31(x+15)=1240

=>x+15=1240:31=40

=>x=40-15=25

1+2+3+...+x=210

=>[(x-1):1+1]. (x+1) : 2= 210

=>x.(x+1):2=210

=>x(x+1)=210.2=420

=>x(x+1)=20.21

=>x=20