a) A = \(\dfrac{\left(x-3\right)\left(x+3\right)-\left(4x-1\right)\left(x-3\right)}{\left(x-3\right)^2}=\dfrac{\left(x-3\right)\left(x+3-4x-1\right)}{\left(x-3\right)^2}=\dfrac{2-3x}{x-3}\)
a) \(A=\dfrac{x^2-9-\left(4x-2\right)\left(x-3\right)}{x^2-6x+9}\left(ĐKXĐ:x\ne3\right)\)
\(\Leftrightarrow\dfrac{\left(x-3\right)\left(x+3\right)-\left(4x-2\right)\left(x-3\right)}{\left(x-3\right)^2}\)
\(\Leftrightarrow\dfrac{\left(x-3\right)\left(x+3-4x+2\right)}{\left(x-3\right)^2}\)
\(\Leftrightarrow\dfrac{\left(x-3\right)\left(-3x+5\right)}{\left(x-3\right)^2}=\dfrac{-3x+5}{x-3}\)
b) Ta có: A = \(\dfrac{-3x+5}{x-3}=\dfrac{-3}{x-3}-4\)
Để A là số nguyên thì \(-3⋮\left(x-3\right)\)
\(\Rightarrow\left(x-3\right)\inƯ\left(-3\right)=\left\{-3;-1;1;3\right\}\)
Do đó:
x - 3 = -3 => x = 0 (nhận)
x - 3 = -1 => x = 2 (nhận)
x - 3 = 1 => x = 4 (nhận)
x - 3 = 3 => x =6 (nhận)
Vậy \(x\in\left\{0;2;4;6\right\}\) thì A nguyên
a/ \(A=\dfrac{x^2-9-\left(4x-2\right)\left(x-3\right)}{x^2-6x+9}\)
\(=\dfrac{\left(x-3\right)\left(x+3\right)-\left(4x-2\right)\left(x-3\right)}{\left(x-3\right)^2}\)
\(=\dfrac{\left(x-3\right)\left[\left(x+3\right)-\left(4x-2\right)\right]}{\left(x-3\right)^2}\)
\(=\dfrac{x+3-4x+2}{x-3}\)
\(=\dfrac{-3x+5}{x-3}\)