\(\left(x-1\right)\left(x+3\right)=6\)

\(\Rightarrow6⋮\left(x-1\right),\left(x+3\right)\)

\(\Rightarrow\left(x-1\right),\left(x+3\right)\inƯ\left(6\right)\)

\(\RightarrowƯ\left(6\right)=\left\{1;2;3;6\right\}\)

Ta có bảng :

Ta co : 8(x-2014)² = 25-y²

=> 8(x-2014)² + y² = 25 ^(*)

Voi moi \(y\in N\) ta co y² \(\ge0\)

\(\Rightarrow8\left(x-2014\right)^2\le25\)

\(\Rightarrow\left(x-2014\right)^2\le\dfrac{25}{3}\)

Vi x\(\in N\)

\(\Rightarrow\left(x-2014\right)^2=0hoac\left(x-2014\right)^2=1\)

Neu\(\left(x-2014\right)^2=1\) thay vao(*) ta duoc;

8 . 1+ y² =25

\(\Rightarrow25-8=y^2\)

17 = y² (loai) (vi y \(\in N\))

Neu \(\left(x-2014\right)^2=0\) thay vao (*) ta duoc:

8 . 0 + y² = 25

=> y² = 25

=> y = 5 (vi y\(\in N\))

Khi do \(\left(x-2014\right)^2=0\)

=> x- 2014 = 0 => x = 2014

Vay x = 2014, y = 5

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.

( x - 4 ) . ( x - 3 ) = 0

Mà một số nhân với 0 = 0

\(\Rightarrow\)1 trong hay vế phải bằng 0

Nếu x = 4

\(\left(4-4\right).\left(4-3\right)=0.1=0\)

Nếu x = 3

\(\left(3-4\right).\left(3-3\right)=-1.0=0\)

\(\Rightarrow x=3;4\)

th1: x- 4= 0                  th2:  x-3= 0

      x= 0+4                         x= 0+3

      x= 4                           x=3

(x-4)*(x-3)=0

X=4hoặc 3

please help me, please!!!!!!!!!!

\(\left(x-1\right)\left(2y+3\right)=1\cdot26=26\cdot1=2\cdot13=13\cdot2\)

Ta co bang sau: