Ta có : \(\left(x-1\right)^{2016}=\left(1-x\right)^{2018}\)

\(\Rightarrow\left(x-1\right)^{2016}=\left(x-1\right)^{2018}\)

\(\Rightarrow x-1=\hept{\begin{cases}0\\1\end{cases}}\).Vì chỉ có 0²⁰¹⁶=0²⁰¹⁸;1²⁰¹⁶=1²⁰¹⁸

\(\Rightarrow x=\hept{\begin{cases}-1\\0\end{cases}}\)

=> x-1=0 hoặc x-1=1

=> x=0 hoặc x=2

Vậy x thuộc {0;2}

Tk mk nha

3/ Ta có:

\(A=\dfrac{1-2x}{x+3}\)

\(A=\dfrac{-2x+1}{x+3}\)

\(A=\dfrac{-2x-6+7}{x+3}\)

\(A=\dfrac{-2\left(x+3\right)+7}{x+3}\)

\(A=-2+\dfrac{7}{x+3}\)

A nguyên khi \(\dfrac{7}{x+3}\) nguyên 

⇒ 7 ⋮ \(x+3\)

\(\Rightarrow x+3\inƯ\left(7\right)\)

\(\Rightarrow x+3\in\left\{1;-1;7;-7\right\}\)

\(\Rightarrow x\in\left\{-2;-4;4;-10\right\}\)

\(\left(x-1\right)^{2016}=\left(1-x\right)^{2018}\)

\(\Rightarrow\left(x-1\right)^{2016}=-\left(x-1\right)^{2018}\)

\(\Rightarrow\left(x-1\right)^{2016}=\left(x-1\right)^{2018}\)

\(\Rightarrow\left(x-1\right)^{2016}-\left(x-1\right)^{2018}=0\)

\(\Rightarrow\left(x-1\right)^{2016}.\left(1-\left(x-1\right)^2\right)=0\)

\(\Rightarrow\orbr{\begin{cases}\left(x-1\right)^{2016}=0\\1-\left(x-1\right)^2=0\end{cases}}\)

nốt nha

\(\left(x-1\right)^{2016}=\left(1-x\right)^{2018}\)

hai vế có mũ là 2016 và 2018 thì đổi ra bằng 0 vì số rất lớn

\(\Rightarrow x\in\left\{1;2\right\}\)

           2x + 3 ⋮ x - 1

⇒ 2x - 2 + 5  ⋮ x - 1

   2(x-1) + 5  ⋮ x - 1

                5  ⋮ x - 1

     x - 1      \(\in\) { -5; -1; 1; 5}

    x            \(\in\) { -4; 0; 2; 6}

a) 6 ⋮ x

⇒ x ∈ Ư(6) = {-6; -3; -2; -1; 1; 2; 3; 6}

b) 8 ⋮ (x + 1)

⇒ x + 1 ∈ Ư(8) ={-8; -4; -2; -1; 1; 2; 4; 8}

⇒ x ∈ {-9; -5; -3; -2; 0; 1; 3; 7}

Vì 6⋮ x

=> x ϵ { 1,2,3,6,-1,-2,-3,-6}

Vì 8 ⋮ x +1

=> x+1 ϵ { 1,2,4,8,-1,-2,-4,-8}

=> x ϵ {0,1,3,7,-2,-3,-5,-9}

=> x = -2