Cứu với mn ơi huhu !!

a) Ta có: \(\dfrac{x}{x-3}-\dfrac{6}{x}-\dfrac{9}{x^2-3x}\)

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

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

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

b) Ta có: \(\dfrac{7}{x}-\dfrac{x}{x+6}+\dfrac{36}{x^2+6x}\)

\(=\dfrac{7\left(x+6\right)-x^2+36}{x\left(x+6\right)}\)

\(=\dfrac{7x+42-x^2+36}{x\left(x+6\right)}\)

\(=\dfrac{-\left(x^2-7x-78\right)}{x\left(x+6\right)}\)

\(=\dfrac{-\left(x^2-13x+6x-78\right)}{x\left(x+6\right)}\)

\(=\dfrac{-\left[x\left(x-13\right)+6\left(x-13\right)\right]}{x\left(x+6\right)}\)

\(=\dfrac{13-x}{x}\)

c) Ta có: \(\dfrac{6}{x-3}-\dfrac{2x-6}{x^2-9}-\dfrac{4}{x+3}\)

\(=\dfrac{6\left(x+3\right)-2x+6-4\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}\)

\(=\dfrac{6x+18-2x+6-4x+12}{\left(x-3\right)\left(x+3\right)}=\dfrac{36}{\left(x-3\right)\left(x+3\right)}\)

ngu ko bit lm nguuuu đúng là cái loại ngu

\(a,\left(3x+x\right)\left(x^2-9\right)-\left(x-3\right)\left(x^2+3x+9\right)\)

\(=4x\left(x^2-9\right)-x^3+27\)

\(=4x^3-36x-x^3+27\)

\(=3x^3-36x+27\)

\(\left(x+6\right)^2-2x.\left(x+6\right)+\left(x-6\right).\left(x+6\right)\)

\(=\left(x+6\right).\left(x+6-2x+x-6\right)\)

\(=\left(x+6\right).0\)

\(=0\)

a,\(\dfrac{x^2-9}{2x+6}:\dfrac{3-x}{2}=\dfrac{\left(x-3\right)\left(x+3\right)}{2\left(x+3\right)}.\dfrac{2}{3-x}=\dfrac{x-3}{3-x}=\dfrac{-\left(3-x\right)}{3-x}=-1\)

b, \(\dfrac{2x}{x-y}-\dfrac{2y}{x-y}=\dfrac{2x-2y}{x-y}=\dfrac{2\left(x-y\right)}{x-y}=2\)

\(a,=\dfrac{\left(x-3\right)\left(x+3\right)}{2\left(x+3\right)}\cdot\dfrac{2}{-\left(x-3\right)}=\dfrac{x-3}{2}\cdot\dfrac{2}{-\left(x-3\right)}=-1\\ b,=\dfrac{2x-2y}{x-y}=\dfrac{2\left(x-y\right)}{\left(x-y\right)}=2\)

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

a: \(\dfrac{x^2}{3x+6}+\dfrac{4x+4}{3x+6}=\dfrac{x^2+4x+4}{3x+6}=\dfrac{x+2}{3}\)

b: \(\dfrac{x+3}{x}+\dfrac{x}{3-x}-\dfrac{9}{3x-x^2}\)

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

=0

a: \(=\dfrac{x+1+3}{x-3}=\dfrac{x+4}{x-3}\)