\(\frac{1}{3x-2}-\frac{1}{3x+2}-\frac{3x-6}{4-9x^2}\)

\(=\frac{3x+2}{9x^2-4}-\frac{3x-2}{9x^2-4}+\frac{3x-6}{9x^2-4}\)

\(=\frac{3x+2-3x+2+3x-6}{9x^2-4}\)

\(=\frac{3x-2}{9x^2-4}\)

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

\(\frac{18}{\left(x-3\right)\left(x^2-9\right)}-\frac{3}{x^2-6x+9}-\frac{x^2}{x^2-9}\)

\(=\frac{18}{\left(x-3\right)\left(x-3\right)\left(x+3\right)}\) \(-\frac{3\left(x+3\right)}{\left(x-3\right)\left(x-3\right)\left(x+3\right)}\)\(-\frac{x^2\left(x-3\right)}{\left(x-3\right)\left(x+3\right)\left(x-3\right)}\)

\(=\frac{18-3x-9-x^3+3x^2}{\left(x-3\right)^2\left(x+3\right)}\)

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

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

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

học tốt

\(ĐKXĐ:\hept{\begin{cases}x\ne0\\x\ne\pm3\end{cases}}\)

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

\(=\left(\frac{9}{x\left(x-3\right)\left(x+3\right)}+\frac{1}{x+3}\right):\left(\frac{x-3}{x\left(x+3\right)}-\frac{x}{3\left(x+3\right)}\right)\)

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

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

\(=\frac{-3}{x-3}\)

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

\(=\frac{-2}{x^2}\)

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

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

\(=x\left(x-3\right)\)

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

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

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

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

\(=\frac{x+3}{x+1}\)

# Sắp ik ngủ nên làm vậy hoi, ko chắc phần kq câu b và c đâu nha

\(\frac{3x}{5x+5y}-\frac{x}{10x-10y}\)

= \(\frac{3x\left(x-y\right)}{5.2.\left(x+y\right)\left(x-y\right)}-\frac{x\left(x+y\right)}{10\left(x^2-y^2\right)}\)

= \(\frac{3x^2-3xy-x^2-xy}{10\left(x^2-y^2\right)}\)

= \(\frac{3x\left(x-y\right)}{10\left(x^2-y^2\right)}\)

= \(\frac{3x}{10\left(x+y\right)}\)