a,\(A=\dfrac{x+4}{x^2-2x}+\dfrac{2}{x}\)
\(=\dfrac{x+4}{x\left(x-2\right)}+\dfrac{2}{x}\)
\(=\dfrac{x+4}{x\left(x-2\right)}+\dfrac{2\left(x-2\right)}{x\left(x-2\right)}\)
\(=\dfrac{x+4+2x-4}{x\left(x-2\right)}\)
\(=\dfrac{3x}{x\left(x-2\right)}=\dfrac{3}{x-2}\)
b, Để A có giá trị bằng - 3
\(\Leftrightarrow\dfrac{3}{x-2}=-3\)
\(\Leftrightarrow x-2=-1\)
\(\Leftrightarrow x=1\) ( t/m )
Vậy x = 1 thì A có giá trị bằng -3