Bài 2:
a) \(A=\left(\dfrac{x+2}{x+1}-\dfrac{x}{x-1}\right).\dfrac{3x+3}{2}.\\ \left(x\ne1;-1\right).\)
\(A=\dfrac{\left(x+2\right)\left(x-1\right)-x\left(x+1\right)}{\left(x+1\right)\left(x-1\right)}.\dfrac{3\left(x+1\right)}{2}.\\ A=\dfrac{x^2-x+2x-2-x^2-x}{x-1}.\dfrac{3}{2}.\\ A=\dfrac{-2}{x-1}.\dfrac{3}{2}=\dfrac{-3}{x-1}.\)
b) \(A< 0.\Leftrightarrow\dfrac{-3}{x-1}< 0.\)
Mà \(-3< 0.\)
\(\Rightarrow x-1>0.\Leftrightarrow x>1.\)
Kết hợp ĐKXĐ.
\(\Rightarrow x>1.\)
Vậy \(A< 0\) thì \(x>1.\)
c) Để \(A\in Z.\Rightarrow\dfrac{-3}{x-1}\in Z.\Leftrightarrow x-1\in\) \(Ư\left(3\right)=\) \(\left\{1;-1;3;-3\right\}.\)
\(\Rightarrow x\in\left\{2;0;4;-2\right\}.\)
Vậy \(A\in Z\) thì \(x\in\left\{2;0;4;-2\right\}.\)
