\(A=\left(\frac{3-x}{x+3}.\frac{x^2+6x+9}{x^2-9}+\frac{x}{x+3}\right):\frac{3x^2}{x+3}\)

\(A=\left(\frac{3-x}{x+3}.\frac{\left(x+3\right)^2}{\left(x-3\right)\left(x+3\right)}+\frac{x}{x+3}\right):\frac{3x^2}{x+3}\)

\(A=\left(\frac{-\left(x-3\right)}{x+3}.\frac{\left(x+3^2\right)}{\left(x-3\right)\left(x+3\right)}+\frac{x}{x+3}\right):\frac{3x^2}{x+3}\)

\(A=\left(-1+\frac{x}{x+3}\right):\frac{3x^2}{x+3}\)

\(A=\left(\frac{-x-3+x}{x+3}\right):\frac{3x^2}{x+3}\)

\(A=\left(-\frac{3}{x+3}\right).\frac{x+3}{3x^2}\)

\(A=-x^2\)

\(ĐKXĐ:x\ne\pm3\)

\(P=\left(\frac{x^2-3x}{x^3+3x^2+9x+27}+\frac{3}{x^2+9}\right):\left(\frac{1}{x-3}-\frac{6x}{x^3-3x^2+9x-27}\right)\)

\(\Leftrightarrow P=\left(\frac{x^2-3x}{\left(x+3\right)\left(x^2+9\right)}+\frac{3}{x^2+9}\right):\left(\frac{1}{x-3}-\frac{6x}{\left(x-3\right)\left(x^2+9\right)}\right)\)

\(\Leftrightarrow P=\frac{\left(x^2-3x\right)+3\left(x+3\right)}{\left(x+3\right)\left(x^2+9\right)}:\frac{x^2+9-6x}{\left(x-3\right)\left(x^2+9\right)}\)

\(\Leftrightarrow P=\frac{x^2+9}{\left(x+3\right)\left(x^2+9\right)}:\frac{\left(x-3\right)^2}{\left(x-3\right)\left(x^2+9\right)}\)

\(\Leftrightarrow P=\frac{1}{x+3}:\frac{x-3}{x^2+9}\)

\(\Leftrightarrow P=\frac{x^2+9}{\left(x+3\right)\left(x-3\right)}\)

\(A=\left(\frac{3-x}{x+3}.\frac{x^2+6x+9}{x^2-9}+\frac{x}{x+3}\right):\frac{3x^2}{x+3}\)

\(=\left[\frac{-\left(x-3\right)}{x+3}.\frac{\left(x+3\right)^2}{\left(x-3\right)\left(x+3\right)}+\frac{x}{x+3}\right].\frac{x+3}{3x^2}\)

\(=\left[\frac{-\left(x-3\right)\left(x+3\right)^2}{\left(x-3\right)\left(x+3\right)^2}+\frac{x}{x+3}\right].\frac{x+3}{3x^2}\)

\(=\left(-1+\frac{x}{x+3}\right).\frac{x+3}{3x^2}\)

\(=\frac{-x-3+x}{x+3}.\frac{x+3}{3x^2}=\frac{-3}{x+3}.\frac{x+3}{3x^2}=\frac{-1}{x^2}\)

b ) Để \(A=-\frac{1}{x^2}< 0\forall x\ne0\)  

Vậy \(x\ne0\) thì \(A< 0\)

M = \(\left(\frac{x}{x-3}-\frac{x+3}{3x^2-6x-9}+\frac{1}{3x+3}\right)\)\(\frac{x^2-2x-3}{x^2+x+2}\)

= \(\left(\frac{x\left(3x+3\right)}{3\left(x-3\right)\left(x+1\right)}-\frac{x+3}{3\left(x-3\right)\left(x+1\right)}+\frac{x-3}{3\left(x+1\right)\left(x-3\right)}\right)\)\(\frac{\left(x+1\right)\left(x-3\right)}{x^2+x+2}\)

=  \(\frac{3\left(x^2+x-2\right)}{3\left(x-3\right)\left(x+1\right)}\)*  \(\frac{\left(x+1\right)\left(x-3\right)}{x^2+x+2}\)  = \(\frac{x^2+x-2}{x^2+x+2}\)

Ta thấy   x² + x - 2  <   x² + x + 2

nên M < 1

ĐK: `x \ne 3; x \ne -3`

`A=3/(x-3)-(6x)/(9-x^2)+x/(x+3)`

`=3/(x-3)+(6x)/(x^2-9)+x/(x+3)`

`=3/(x-3)+(6x)/((x-3)(x+3))+x/(x+3)`

`=(3(x+3)+6x+x(x-3))/((x-3)(x+3))`

`=(3x+9+6x+x^2-3x)/((x+3)(x-3))`

`=(x^2+6x+9)/((x-3)(x+3))`

`=((x+3)^2)/((x-3)(x+3))`

`=(x+3)/(x-3)`

`x=5 => A=(5+3)/(5-3)=4`

ĐKXĐ:\(\left\{{}\begin{matrix}x-3\ne0\\9-x^2\ne0\\x+3\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne3\\x^2\ne9\\x\ne-3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne3\\x\ne-3\end{matrix}\right.\)

\(\dfrac{3}{x-3}-\dfrac{6x}{9-x^2}+\dfrac{x}{x+3}\\ =\dfrac{3\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}-\dfrac{6x}{\left(3-x\right)\left(3+x\right)}+\dfrac{x\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}\\ =\dfrac{3x+9}{\left(x-3\right)\left(x+3\right)}+\dfrac{6x}{\left(x-3\right)\left(x+3\right)}+\dfrac{x^2-3x}{\left(x-3\right)\left(x+3\right)}\\ =\dfrac{3x+9+6x+x^2-3x}{\left(x-3\right)\left(x+3\right)}\\ =\dfrac{x^2+6x+9}{\left(x-3\right)\left(x+3\right)}\\ =\dfrac{\left(x+3\right)^2}{\left(x-3\right)\left(x+3\right)}\\ =\dfrac{x+3}{x-3}\)

Thay x=5 vào \(\dfrac{x+3}{x-3}=\dfrac{5+3}{5-3}=\dfrac{8}{2}=4\)