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干天杭3927已知a+b+c=0,求a(1/b+1/c)+b(1/a+1/c)+c(1/a+1/b)的值我们老师说是 - 3,可我答得是0,老师还没讲这道题,请高手帮下忙. -
幸炊罗13987702843 ______[答案] a(1/b+1/c)+b(1/c+1/a)+c(1/a+1/b) = c/b + a/b + a/c + b/c + b/a + c/a = a+c/b + a+b/c + b+c/a 因为a+b=-c a+c=-b b+c=-a 原式(-1)+ (-1) + (-1) =-3

干天杭3927已知:abc=1求,a/1+a+ab+b/1+b+bc+c/1+c+ca= -
幸炊罗13987702843 ______[答案] 题意不明,不过根据我的猜测应该是这样的 因为abc=1. a/(ab+a+1)+b/(bc+b+1)+c/(ac+c+1) =a/(ab+a+1)+ab/(abc+ab+a)+abc/(abac+abc+ab) =a/(ab+a+1)+ab/(1+ab+a)+1/(a+1+ab) =(ab+a+1)/(ab+a+1) =1

干天杭3927已知A+B+C=61,A+C+D=71,A+B+D=62,B+C+D=64.求A.B.C.D各是多少? -
幸炊罗13987702843 ______[答案] 已知A+B+C=61,A+C+D=71,A+B+D=62,B+C+D=64. 所以:四项相加,3(A+B+C+D)=258. (A+B+C+D)=86 A=22 B=15 C=24 D=25

干天杭3927因式分解abc+ab+bc+ca+a+b+c+1= -
幸炊罗13987702843 ______[答案] abc+ab+bc+ca+a+b+c+1 =(abc+ab)+(bc+b)+(ca+a)+(c+1) =ab(c+1)+b(c+1)+a(c+1)+(c+1) =(c+1)(ab+a+b+1) =(c+1)[(ab+a)+(b+1)] =(c+1)[a(b+1)+(b+1)] =(c+1)(b+1)(a+1)

干天杭3927已知a+b+c=1,求证(1/a)+(1/b)+(1/c)≥9 -
幸炊罗13987702843 ______[答案] 证明:∵a+b+c=1∴1/a+1/b+1/c=(a+b+c)/a+(a+b+c)/b+(a+b+c)/c=1+b/a+c/a+a/b+1+c/b+a/c+b/c+1=3+(b/a+a/b)+(c/a+a/c)+(b/c+c/b)≥3+2+2+2=9 (b/a+a/b≥2,c/a+a/c...

干天杭3927已知abc=1,求(ab+a+1)分之a+(bc+b+1)分之b+(ac+c+1)分之c -
幸炊罗13987702843 ______[答案] abc=1 a/(ab+a+1)=ac/(abc+ac+c)=ac/(ac+c+1) b/(bc+b+1)=abc/(abc^2+abc+ac)=1/(ac+c+1) (ab+a+1)分之a+(bc+b+1)分之b+(ac+c+1)分之c =ac/(ac+c+1)+1/(ac+c+1)+c/(ac+c+1) =(ac+1+c)/(ac+c+1) =1

干天杭3927逻辑式 F = A+B 可变换为 - 上学吧普法考试
幸炊罗13987702843 ______ Is it +adj +for sb+to do?改成一般形式是it is +adj +for sb+to do 比如 is it easy for me to read this book?改成一般形式是 it is easy for me to read this book

干天杭3927已知a,b,c满足ab+a+b=ac+c+a=3,求(a+1)(b+1)(c+1)的值 -
幸炊罗13987702843 ______[答案] ab+a+b=bc+b+c=ac+a+c=3,则ab+a+b+1=bc+b+c+1=ac+a+c+1=4于是,a(b+1)+b+1=(a+1)(b+1)=4,同理得:(b+1)(c+1)=4,(c+1)(a+1)=4三式相乘得:[(a+1)(b+1)(c+1)]²=4*4*4=64所以:(a+1)(b+1)(c+1)=8...