\(A=\frac{x^3-4x^2+4x-10}{x-3}\)( ĐKXĐ : x ≠ 3 )
\(=\frac{x^3-3x^2-x^2+3x+x-3-7}{x-3}\)
\(=\frac{x^2\left(x-3\right)-x\left(x-3\right)+\left(x-3\right)-7}{x-3}\)
\(=\frac{\left(x-3\right)\left(x^2-x+1\right)-7}{x-3}\)
\(=\frac{\left(x-3\right)\left(x^2-x+1\right)}{x-3}-\frac{7}{x-3}\)
\(=\left(x^2-x+1\right)-\frac{7}{x-3}\)
Vì x ∈ Z nên ( x2 - x + 1 ) ∈ Z
nên để A ∈ Z thì \(\frac{7}{x-3}\)∈ Z
hay ( x - 3 ) ∈ Ư(7) = { ±1 ; ±7 }
| x-3 | 1 | -1 | 7 | -7 |
| x | 4 | 2 | 10 | -4 |
Các giá trị tm ĐKXĐ
Vậy x ∈ { ±4 ; 2 ; 10 } thì A ∈ Z
\(ĐKXĐ:x\ne3\)
\(A=\frac{x^3-4x^2+4x-10}{x-3}=\frac{x^3-3x^2-x^2+3x+x-3-7}{x-3}\)
\(=\frac{x^2\left(x-3\right)-x\left(x-3\right)+\left(x-3\right)-7}{x-3}\)
\(=\frac{\left(x-3\right)\left(x^2-x+1\right)-7}{x-3}=\left(x^2-x+1\right)-\frac{7}{x-3}\)
Vì \(x\inℤ\)\(\Rightarrow x^2-x+1\inℤ\)
\(\Rightarrow\)Để \(A\inℤ\)thì \(\frac{7}{x-3}\inℤ\)\(\Rightarrow7⋮x-3\)
\(\Rightarrow x-3\inƯ\left(7\right)=\left\{-7;-1;1;7\right\}\)
\(\Leftrightarrow x\in\left\{-4;2;4;10\right\}\)( thỏa mãn ĐKXĐ )
Vậy \(x\in\left\{-4;2;4;10\right\}\)
