Question

Решите пожалуйста!!!!​

Answer 1

$1) \ \dfrac{x-9}{x^{2}-9 }-\dfrac{3}{3x-x^{2} } =\dfrac{x-9}{(x-3)(x+3)}-\dfrac{3}{x(3-x)}=\dfrac{x-9}{(x-3)(x+3)}+\dfrac{3}{x(x-3)}=\\\\\\=\dfrac{x\cdot(x-9)+3\cdot(x+3)}{x(x-3)(x+3)}=\dfrac{x^{2}-9x+3x+9 }{x(x-3)(x+3)}=\dfrac{x^{2}-6x+9 }{x(x-3)(x+3)}=\\\\\\=\dfrac{(x-3)^{2} }{x(x-3)(x+3)} =\boxed{\dfrac{x-3}{x^{2}+3x }}$

$2) \ \dfrac{1}{x-2}-\dfrac{6x}{x^{3}-8 } =\dfrac{1}{x-2}-\dfrac{6x}{(x-2)(x^{2}+2x+4) }=\\\\\\=\dfrac{x^{2}+2x+4-6x }{(x-2)(x^{2}+2x+4) }=\dfrac{x^{2}-4x+4 }{(x-2)(x^{2}+2x+4) }=\dfrac{(x-2)^{2} }{(x-2)(x^{2}+2x+4) } =\\\\\\=\boxed{\dfrac{x-2}{x^{2}+2x+4 } }$

$3) \ \dfrac{3}{x} +\dfrac{21}{x^{2}-7x } -\dfrac{4-x}{x-7} = \dfrac{3}{x} +\dfrac{21}{x(x-7)} -\dfrac{4-x}{x-7}=\dfrac{3\cdot(x-7)+21-(4-x)\cdot x}{x(x-7)} =\\\\\\=\dfrac{3x-21+21-4x+x^{2} }{x(x-7)} =\dfrac{x^{2}- x}{x(x-7)} =\dfrac{x(x-1)}{x(x-7)} =\boxed{\dfrac{x-1}{x-7}}$