a.\(A=2+\dfrac{8x\left(x+3\right)}{\left(x-3\right)^2}:\left(\dfrac{x+3}{x}+\dfrac{2}{x-3}+\dfrac{15-x^2}{x\left(x-3\right)}\right)\)
\(A=2+\dfrac{8x\left(x+3\right)}{\left(x-3\right)^2}:\left(\dfrac{\left(x+3\right)\left(x-3\right)+2x+15-x^2}{x\left(x-3\right)}\right)\)
\(A=2+\dfrac{8x\left(x+3\right)}{\left(x-3\right)^2}:\left(\dfrac{x^2-9+2x+15-x^2}{x\left(x-3\right)}\right)\)
\(A=2+\dfrac{8x\left(x+3\right)}{\left(x-3\right)^2}.\dfrac{x\left(x-3\right)}{2x+6}\)
\(A=2+\dfrac{8x\left(x+3\right)x\left(x-3\right)}{\left(x-3\right)^22\left(x+3\right)}\)
\(A=2+\dfrac{4x^2}{x-3}\)
b.Để A>0 thì \(x-3>0\) ( vì \(4x^2>0\) )
\(\Leftrightarrow x>3\)
c.\(\left|A\right|=3-3x\) ; \(ĐK:x\le1\)
\(\Leftrightarrow\left|2+\dfrac{4x^2}{x-3}\right|=3-3x\)
\(\Leftrightarrow\left[{}\begin{matrix}2+\dfrac{4x^2}{x-3}=3-3x\\-2-\dfrac{4x^2}{x-3}=3-3x\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\dfrac{4x^2}{x-3}=1-3x\\-\dfrac{4x^2}{x-3}=5-3x\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}4x^2=\left(x-3\right)\left(1-3x\right)\\-4x^2=\left(5-3x\right)\left(x-3\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}4x^2=x-3x^2-3+9x\\-4x^2=5x-15-3x^2+9\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}7x^2-10x+3=0\\-x^2-5x+6=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x=1\left(l\right)\\x=\dfrac{3}{7}\left(n\right)\end{matrix}\right.\\\left[{}\begin{matrix}x=1\left(l\right)\\x=-6\left(n\right)\end{matrix}\right.\end{matrix}\right.\)
Vậy \(x=\left\{\dfrac{3}{7};-6\right\}\)
