已知△ABC的三边分别a b c且满足a2+b2+c2+338=10a+24b+26c.试判断△ABC的形状

a^2+b^2+c^2+388=10a+24b+26c.所以:(a^2-5a+25)+(b^2-24a+144)+(c^2-26a+169)=0.所以:(a-5)^2+(b-12)^2+(c-13)^2=0.所以a=5,b=12,c=13.所以:a^2+b^2=c^2,根据勾股定理的逆定理,这是直角三角形....