\(P=\frac{x^3-6x^2+11x-12}{x^2-5x+4}\)
\(=\frac{\left(x^3-4x^2\right)-\left(2x^2-8x\right)+\left(3x-12\right)}{\left(x^2-4x\right)-\left(x-4\right)}\)
\(=\frac{x^2\left(x-4\right)-2x\left(x-4\right)+3\left(x-4\right)}{x\left(x-4\right)-\left(x-4\right)}\)
\(=\frac{\left(x-4\right)\left(x^2-2x+3\right)}{\left(x-4\right)\left(x-1\right)}\)
\(=\frac{x^2-2x+3}{x-1}\)
Để P nguyên thì \(\frac{x^2-2x+3}{x-1}\) nguyên
\(\Rightarrow x^2-2x+3⋮x-1\)
\(\Rightarrow\left(x-1\right)^2+2⋮x-1\)
\(\Leftrightarrow x-1\in\left\{1;2;-1;-2\right\}\)
\(\Leftrightarrow x\in\left\{2;3;0;-1\right\}\)
a) \(P=\frac{x^3-6x^2+11x-12}{x^2-5x+4}\)
\(=\frac{x^3-4x^2-2x^2+3x+8x-12}{x^2-4x-x+4}\)
\(=\frac{\left(x^3-4x^2\right)-\left(2x^2-8x\right)+\left(3x-12\right)}{x^2-4x-x+4}\)
\(=\frac{x^2\left(x-4\right)-2x\left(x-4\right)+3\left(x-4\right)}{x\left(x-4\right)-\left(x-4\right)}\)
\(=\frac{\left(x^2-2x+3\right)\left(x-4\right)}{\left(x-1\right)\left(x-4\right)}\)
\(=\frac{x^2-2x+3}{x-1}\)
b)
Suy ra để P nguyên thì \(x-1\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\)
| \(x-1\) | \(1\) | \(-1\) | \(2\) | \(-2\) |
| \(x\) | \(2\) | \(0\) | \(3\) | \(-1\) |
