作业帮若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+cd+c+1)+(d/dab+da+d+1)的值
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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+cd+c+1)+(d/dab+da+d+1)的值
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+cd+c+1)+(d/dab+da+d+1)的值
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+cd+c+1)+(d/dab+da+d+1)的值
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
21.5