Răspuns
[tex]Din~proprietatiile~echivalente~stim~ca~ \dfrac{a}{x}=\dfrac{b}{y}=\dfrac{c}{z}=\dfrac{a+b+c}{x+y+z}\\deci~\dfrac{a+b+c}{x+y+z}=\dfrac{3}{11}\\Prin~urmare~ 11\cdot \dfrac{a+b+c}{x+y+z}-3=11\cdot \dfrac{3}{11}-3=3-3=\bold{0}[/tex]
Explicație pas cu pas:
