Lời giải:
ĐKXĐ: \(x\neq \pm 3\)
a)
Ta có \(P=\frac{2x(3-x)}{(x+3)(3-x)}-\frac{x(x+3)}{(x+3)(3-x)}+\frac{6x}{(3-x)(3+x)}\)
\(=\frac{2x(3-x)-x(x+3)+6x}{(3-x)(3+x)}=\frac{9x-3x^2}{(3-x)(3+x)}=\frac{3x(3-x)}{(3-x)(3+x)}=\frac{3x}{3+x}\)
b) Tại $x=\frac{3}{4}$
\(\Rightarrow P=\frac{3.\frac{3}{4}}{3+\frac{3}{4}}=\frac{3}{5}\)