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.
来源:学生作业帮助网 编辑:作业帮 时间:2024/06/27 16:06:43
![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.](/uploads/image/z/2639504-56-4.jpg?t=%E8%AE%BEabcd%3D1%2C%E6%B1%82%E8%AF%81%EF%BC%9A%28a%2Fabc%2Bab%2Ba%2B1%29%2B%28b%2Fbcd%2Bbc%2Bb%2B1%29%2B%28c%2Fcda%2Bcd%2Bc%2B1%29%2B%28d%2Fabd%2Bad%2Bd%2B1%29%3D1.)
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.
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.
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、、、
等式左边乘以abcd,后合并同类项即可证得(a/abc+ab+a+1)+(b/bcd+bc+b+1)+(c/cda+cd+c+1)+(d/abd+ad+d+1)=1.