Question

已知a,b,c满足a^-6b=-15,b^-8c=-19,c^-4a=5a,求a+b+c的值．

Answer 1

已知a,b,c满足a^-6b=-15,b^-8c=-19,c^-4a=5,求a+b+c的值.

将三个等式相加,得

a^2-6b+b^2-8c+c^2-4a=-29

(a^2-4a+4)+(b^2-6b+9)+(c^2-8c+16)=0

(a-2)^2+(b-3)^2+(c-4)^2=0

上述三个数的平方和为0,则此三个数都为0,即

(a-2)^2=0,解得：a=2,

(b-3)^2=0,解得：b=3,

(c-4)^2=0,解得：c=4,

所以

a+b+c=2+3+4=9.