当abcd=1时,(a/abc+ab+a+1)+(b/bcd+bc+b+1)+(c/cda+cd+c+1)+(d/bad+bd+d+1)

Question

如题所述

Answer 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