Question

(a+b+c)/(a+b-c)=7, (a+b+c)/(a+c-b)=3,5, чему равно (a+b+c)/(b+c-a)

Answer 1

Ответ:

$\dfrac{a+b+c}{a+b-c}=7\ \ ,\ \ \dfrac{a+b+c}{a+c-b}=3,5\ \ ,\ \ \ \dfrac{a+b+c}{b+c-a}=?\\\\\\\dfrac{a+b+c}{a+b-c}=7=\dfrac{7}{1}\ \ ,\ \ \dfrac{a+b+c}{a+c-b}=3,5=\dfrac{7}{2}\ \ \to \ \ \left\{\begin{array}{l}a+b+c=7\\a+b-c=1\\a+c-b=2\end{array}\right\\\\\\1\ yravnenie-2\ yravnenie\ ;\ \ 1\ yravnenie-3\ yravnenie\ ;$

$\left\{\begin{array}{l}a+b+c=7\\2c=7-1\\2b=7-2\end{array}\right\ \ \ \left\{\begin{array}{l}a=7-b-c\\2c=6\\2b=5\end{array}\right\ \ \ \left\{\begin{array}{l}a=7-2,5-3\\c=3\\b=2,5\end{array}\right\ \ \left\{\begin{array}{l}a=1,5\\b=2,5\\c=3\end{array}\right$

$\dfrac{a+b+c}{b+c-a}=\dfrac{1,5+2,5+3}{2,5+3-1,5}=\dfrac{7}{4}=1,75$