\(A=\dfrac{x+2}{x+3}-\dfrac{5}{\left(x-2\right)\left(x+3\right)}+\dfrac{1}{2-x}\)
a ) ĐKXĐ :\(x\ne2\) và \(x\ne-3\).
Rút gọn : \(A=\dfrac{x+2}{x+3}-\dfrac{5}{\left(x-2\right)\left(x+3\right)}+\dfrac{1}{2-x}\)
\(\Leftrightarrow A=\dfrac{x+2}{x+3}-\dfrac{5}{\left(x-2\right)\left(x+3\right)}-\dfrac{1}{x-2}\)
\(\Leftrightarrow A=\dfrac{\left(x+2\right)\left(x-2\right)-5-x-3}{\left(x+3\right)\left(x-2\right)}\)
\(\Leftrightarrow A=\dfrac{x^2-x-12}{\left(x+3\right)\left(x-2\right)}=\dfrac{\left(x-4\right)\left(x+3\right)}{\left(x+3\right)\left(x-2\right)}=\dfrac{x-4}{x-2}\)
b ) Khi \(A=-\dfrac{3}{4},\) thì :
\(\dfrac{x-4}{x-2}=-\dfrac{3}{4}\)
\(\Leftrightarrow4\left(x-4\right)=-3\left(x-2\right)\)
\(\Leftrightarrow4x-16=-3x+6\)
\(\Leftrightarrow x=\dfrac{22}{7}\).
c ) Ta có : \(\dfrac{x-4}{x-2}=\dfrac{x-2-2}{x-2}=1-\dfrac{2}{x-2}\)
Vậy để A nguyên thi \(x-2⋮2\inƯ\left(2\right)=\left\{1;-1;2;-2\right\}\)
Thay vào từng cái sẽ ra nha :**
d ) Ta có : \(x^2-9=0\)
\(\Leftrightarrow\left(x-3\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\)
+ ) Khi x = 3 , thì :
\(A=\dfrac{3-4}{3-2}=\dfrac{-1}{1}=-1\)
+ ) Khi x = -3, thì :
\(A=\dfrac{-3-4}{-3-2}=\dfrac{-7}{-5}=\dfrac{7}{5}.\)
Vậy ........
