已知a,b,c是三角形ABC的三条边,方程(b+c)x平方+根号2(a-c)x-3/4(a-c)=0有两个相等的实数根,试说明三角形ab

(b+c)x^2+[2(a-c)]^0.5*x-3(a-c)/4=0有两个相等的实数根：[2^0.5(a-c)]^2-4*(b+c)*[3(a-c)/4]=02(a-c)^2-3(b+c)(a-c)=0(a-c)[2(a-c)-3(b+c)]=0a=c;2a-3b-5c=0a=(3b+5c)/2=3(b+c)/2+c>b+c,不成立所以,a=c三角形ABC是...