Question

Упростите выражение $(\frac{x}{x-3} - \frac{2}{x+3} ) : \frac{4x^{2}+4x+24 }{x^{2}-9 }$
Докажите тождество $\frac{9-p^{2} }{3p+9} . \frac{p^{2} }{(3-p)^{2} } +\frac{p}{p-3} = -\frac{p}{3}$
Заранее спасибо)

Answer 1

$1)(\frac{x}{x-3}-\frac{2}{x+3}):\frac{4x^{2}+4x+24 }{x^{2}-9 }=\frac{x*(x+3)-2*(x-3)}{(x-3)(x+3)}*\frac{(x-3)(x+3)}{4(x^{2}+x+6)}=\\\\=\frac{x^{2}+3x-2x+6}{(x-3)(x+3)} *\frac{(x-3)(x+3)}{4(x^{2}+x+6)}=\frac{x^{2}+x+6}{(x-3)(x+3)} *\frac{(x-3)(x+3)}{4(x^{2}+x+6)} =\frac{1}{4}=0,25$

$2)\frac{9-p^{2}}{3p+9}*\frac{p^{2}}{(3-p)^{2}}+\frac{p}{p-3}=\frac{(3-p)(3+p)}{3(p+3)} *\frac{p^{2}}{(3-p)^{2}}+\frac{p}{p-3}=\frac{p^{2}}{3(3-p)}-\frac{p}{3-p}=\\\\=\frac{p^{2}-3p }{3(3-p)}=\frac{p(p-3)}{3(3-p)}=-\frac{p(3-p)}{3(3-p)}=-\frac{p}{3}\\\\-\frac{p}{3}=-\frac{p}{3}$

Доказано