设abcd=1,求证：(a/abc+ab+a+1)+(b/bcd+bc+b+1)+(c/cda+cd+c+1)+(d/a

来源：学生作业帮编辑：拍题作业网作业帮分类：数学作业时间：2024/04/24 16:55:27

设abcd=1,求证：(a/abc+ab+a+1)+(b/bcd+bc+b+1)+(c/cda+cd+c+1)+(d/abd+ad+d+1)=1.

a/(abc+ab+a+1)+b/(bcd+bc+b+1)+c/(cda+cd+c+1)+d/(dab+da+d+1) =a/(1/d+ab+a+1)+b/(bcd+bc+b+1)+c/(1/b+cd+c+1)+d/(dab+da+d+1) =ad/(abd+ad+d+1)+b/(bcd+bc+b+1)+bc/(bcd+bc+b+1)+d/(dab+da+d+1) =(ad+d)/(abd+ad+d+1)+(b+bc)/(bcd+bc+b+1) =(ad+d)/(abd+ad+d+abcd)+(b+bc)/(bcd+bc+b+abcd) =(a+1)/(ab+a+1+abc)+(1+c)/(cd+c+1+acd) =(a+1)/[(a+1)+ab(c+1)]+(c+1)/[(c+1)+cd(a+1)] =1/[1+ab(c+1)/(a+1)]+1/[1+cd(a+1)/(c+1)] =1/{1+(c+1)/[cd(a+1)]}+1/[1+cd(a+1)/(c+1)] 令(c+1)/[cd(a+1)]=x 则cd(a+1)/(c+1)=1/x 所以原式=1/(1+x)+1/(1+1/x) =1/(1+x)+x/(1+x) =(1+x)/(1+x) =1
给分哈、、、

分式设abcd=1,则a/(abc+ab+a+1)+b/(bcd+bc+b+1)+c/(cda+cd+c+1)+d/(d 已知abcd=1,求a/(abc+ab+a+1)+b/(bcd+bc+b+1)+c/(cda+cd+c+1)+d/(da 若a、b、c、d是四个正数,且abcd=1.求(a/abc+ab+a+1)+(b/bcd+bc+b+1)+(c/cda+ 证明(abc+bcd+cda+dab)^2-(ab-cd)(bc-da)(ca-bd)=abcd(a+b+c+d)^2 已知abcd=1求1/(abc+ab+a+1)+1/(bcd+bc+b+1)+1/(cda+cd+c+1)+1/(dab 已知abcd=1,求ax/(abc+ab+a+1)+bx/(bcd+bc+b+1)+cx/(cda+cd+c+1)+dx abcd+abc+bcd+cda+dab+ab+bc+cd+da+ac+bd+a+b+c+d=2009 a+b+c+d= abcd=1 求1+a+ab+abc分之1 + 1+b+bc+bcd分之1 + 1+d+cd+cda分之1 + 1+d+ 设a,b,c满足ab+bc+cd+da=1,求证：a^3/(b+c+d)+b^3/(a+c+d)+c^3/(a+b+d) abcd是实数,ad-bc=1,求证：a+b+c+d+ab+cd≠1 abcd+abc+abd+acd+bcd+ab+ac+ad+bc+bd+cd+a+b+c+d=2009,a+b+c+d= 1,设a.b.c都是正数,求证：（ab+cd）（ac+bd)≥4abcd