09/07/12 16:10:30
>>152
題意から
a/(a+2) + b/(b+2) + c/(c+2) = 1,
そこで ボクは
x = k * a/(a+2),
y = k * b/(b+2),
z = k * c/(c+2),
とおいた。
x+y+z = k > 0,
a = 2x/(k-x) = 2x/(y+z),
b = 2y/(k-y) = 2y/(z+x),
c = 2z/(k-z) = 2z/(x+y),
09/07/12 16:10:30
>>152
題意から
a/(a+2) + b/(b+2) + c/(c+2) = 1,
そこで ボクは
x = k * a/(a+2),
y = k * b/(b+2),
z = k * c/(c+2),
とおいた。
x+y+z = k > 0,
a = 2x/(k-x) = 2x/(y+z),
b = 2y/(k-y) = 2y/(z+x),
c = 2z/(k-z) = 2z/(x+y),