Question

9a2-12ab-4bc+2c2-4c+4=0求a b c 的值

Answer 1

原式=(9a^2-12ab+4b^2)+(4b^2-4bc+c^2)+(c^2-4c+4)=0

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

(3a-2b)^2>=0 (2b-c)^2>=0 (c-2)^2>=0

所以3a-2b=0 2b-c=0 c-2=0

c=2,b=1,a=2/3