\(B=\frac{x^2-4x+5}{x-2}=\frac{x^2-2.x.2+2^2+1}{x-2}=\frac{\left(x-2\right)^2+1}{x-2}\)
\(=\frac{\left(x-2\right)^2}{x-2}+\frac{1}{x-2}=x-2+\frac{1}{x-2}\)
Để B nguyên thì \(\frac{1}{x-2}\) nguyên
=>1 chia hết chi x-2
=>x-2\(\inƯ\left(1\right)\)
=>x-2\(\in\left\{-1;1\right\}\)
=>x-2\(\in\left\{1;3\right\}\)
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