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

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

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

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

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

\(\Rightarrow dpcm\)

TK NHA !!! Vì ko có thời gian nên làm hơi tắt !!!

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

học tốt

\(ĐKXĐ:x\ne\pm3\)

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

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

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

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

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

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

= \(\left[\frac{x.\left(x+3\right)}{\left(x+3\right).\left(x^2+9\right)}+\frac{3}{x+9}\right]:\left[\frac{1}{x-3}-\frac{6x}{\left(x-3\right)\left(x^2+9\right)}\right]\) ]

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

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

= \(\frac{x+3}{x-3}\)

k mik nhé. Plssss~

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