\(a.\)
\(C=\dfrac{2x-9}{x^2-5x+6}-\dfrac{x^2+3x}{x^2-2x}-\dfrac{2x+1}{3-x}\)
\(C=\dfrac{2x-9}{\left(x-3\right)\left(x-2\right)}-\dfrac{x^2+3x}{x\left(x-2\right)}+\dfrac{2x+1}{x-3}\) \(\left(1\right)\)
\(\text{Đ}KX\text{Đ}:\) \(\left\{{}\begin{matrix}x\ne0\\x\ne2\\x\ne3\end{matrix}\right.\)
\(\left(1\right)\Rightarrow\) \(C=\dfrac{2x-9}{\left(x-3\right)\left(x-2\right)}-\dfrac{x^2+3x}{x\left(x-2\right)}+\dfrac{2x+1}{x-3}\)
\(C=\dfrac{2x-9}{\left(x-3\right)\left(x-2\right)}-\dfrac{x+3}{x-2}+\dfrac{2x+1}{x-3}\)
\(C=\dfrac{2x-9}{\left(x-3\right)\left(x-2\right)}-\dfrac{\left(x+3\right)\left(x-3\right)}{\left(x-2\right)\left(x-3\right)}+\dfrac{\left(2x+1\right)\left(x-2\right)}{\left(x-3\right)\left(x-2\right)}\)
\(C=\dfrac{2x-9-x^2+9+2x^2-3x-2}{\left(x-3\right)\left(x-2\right)}\)
\(C=\dfrac{x^2-x-2}{\left(x-3\right)\left(x-2\right)}\)
\(C=\dfrac{\left(x-2\right)\left(x-1\right)}{\left(x-3\right)\left(x-2\right)}\)
\(C=\dfrac{x-1}{x-3}\)
\(b\)
\(C=\dfrac{x-1}{x-3}=\dfrac{\left(x-3\right)+4}{x-3}=1+\dfrac{4}{x-3}\)
Để C nguyên thì \(x-3\in\text{Ư}\left(4\right)=\left\{-4;-2;-1;1;2;4\right\}\)
\(\Rightarrow x\in\left\{-1;1;2;4;5;7\right\}\)
\(a.C=\dfrac{2x-9}{x^2-5x+6}-\dfrac{x^2+3x}{x^2-2x}-\dfrac{2x+1}{3-x}\) ( x # 2 ; x # 0 ; x # 3 )
\(C=\dfrac{2x^2-9x}{x\left(x-2\right)\left(x-3\right)}-\dfrac{x\left(x^2-9\right)}{x\left(x-2\right)\left(x-3\right)}+\dfrac{\left(x^2-2x\right)\left(2x+1\right)}{x\left(x-2\right)\left(x-3\right)}\) \(C=\dfrac{2x^2-9x-x^3+9x+2x^3-3x^2-2x}{x\left(x-2\right)\left(x-3\right)}\)
\(C=\dfrac{x^3-x^2-2x}{x\left(x-2\right)\left(x-3\right)}\)
\(C=\dfrac{x\left(x-2\right)\left(x+1\right)}{x\left(x-2\right)\left(x-3\right)}=\dfrac{x+1}{x-3}\)
b. \(C=\dfrac{x+1}{x-3}=\dfrac{x-3+4}{x-3}=1+\dfrac{4}{x-3}\)
Để : C ∈ Z ⇒ ( x - 3 )∈ { 1 ; -1 ; 2 ; -2 ; 4 ; -4 }
Vậy ,....
