Question

Умоляю вас иначе мне будет

Answer 1

$\dfrac{a+b+c}{a+b-c}=3\ \ ,\ \ \dfrac{a+b+c}{a+c-b}=4,5\ \ ,\ \ \ \dfrac{a+b+c}{b+c-a}=?\\\\\\\dfrac{a+b+c}{a+b-c}=3=\dfrac{3}{1}=\dfrac{9}{3}\ \ ,\ \ \dfrac{a+b+c}{a+c-b}=4,5=\dfrac{9}{2}\ \ \to \ \ \left\{\begin{array}{l}a+b+c=9\\a+b-c=3\\a+c-b=2\end{array}\right\\\\\\1\ yravnenie-2\ yravnenie\ \ ;\ \ 1\ yravnenie-3\ yravnenie\ \ ;$

$\left\{\begin{array}{l}a+b+c=9\\2c=9-3\\2b=9-2\end{array}\right\ \ \ \left\{\begin{array}{l}a=9-b-c\\2c=6\\2b=7\end{array}\right\ \ \ \left\{\begin{array}{l}a=9-3,5-3\\c=3\\b=3,5\end{array}\right\ \ \left\{\begin{array}{l}a=2,5\\b=3,5\\c=3\end{array}\right$

$\dfrac{a+b+c}{b+c-a}=\dfrac{2,5+3,5+3}{3,5+3-2,5}=\dfrac{9}{4}=2,25$