T=[(b/a)+(d/c)]

T=[(b/a)+(d/c)] int a,b,c,t=0; scanf("%d%d%d",&a,&b,&c); if(a>b) { t=a; a=b; t=MAX（a+b,c+d）*10 设集合A={a,b,c,d},A上的二元关系R={(a,b)(b,a)(b,c)(c,d)}求t(R) ( )-(c-d)=(a-c)-(-b+d) # include void main() {int a,b,c,t; scanf("%d%d%d",&a,&b,&c) a b c d* d_________=d c b a 实数a,b,c,d满足d>c；a+b=c+d；a+d 已知a:b=c:d,求证(a+c):(a-c)=(b+d):(b-d) a>b>c>d>0.a/b=c/d怎么证明a+d>c+b D = b+c c+d a+b 求证(b,c,d)a+(c,a,d)b+(a,b,d)c+(b,a,c)d=0 a,b,c,d皆为向量>