Question

Tìm x thuộc z để A có giá trị nguyên

A=\(\frac{4x}{x-2}\)

Answer 1

\(A=\frac{4x}{x-2}=\frac{4\left(x-2\right)+8}{x-2}=4+\frac{8}{x-2}\)

Để A có giá trị nguyên thì \(x-2\inƯ\left(8\right)=\left\{\pm1;\pm2;\pm4;\pm8\right\}\)

Xét bảng ( bạn tự xét nha )

KL

Answer 2

We have \(A=\frac{4x}{x-2}=\frac{4\left(x-2\right)+8}{x-2}=4+\frac{8}{x-2}\)

\(A\inℤ\Leftrightarrow\frac{8}{x-2}\inℤ\Leftrightarrow8⋮\left(x-2\right)\)

\(\Rightarrow x-2\inƯ\left(8\right)=\left\{\pm1;\pm2;\pm4;\pm8\right\}\)

Prints:

\(x-2\)	\(1\)	\(-1\)	\(2\)	\(-2\)	\(4\)	\(-4\)	\(8\)	\(-8\)
\(x\)	\(3\)	\(1\)	\(4\)	\(0\)	\(6\)	\(-2\)	\(10\)	\(-6\)

So \(x\in\left\{3;1;4;0;6;-2;10;-6\right\}\)

Answer 3

Để A có giá trị nguyên thì: \(4x⋮x-2\)

\(\Rightarrow4x-4\left(x-2\right)⋮x-2\)

\(\Rightarrow4x-4x+8⋮x-2\)

\(\Rightarrow8⋮x-2\Rightarrow x-2\inƯ\left(8\right)\)

Mà \(Ư\left(8\right)=\left\{-1;-2;-4;-8;8;4;2;1\right\}\)

\(\Rightarrow n-2\in\left\{-1;-2;-4;-8;8;4;2;1\right\}\)

\(\Rightarrow n\in\left\{1;0;-2;-6;10;6;4;3\right\}\)