提问:若abc=1,求1/(ab+a+1) 1/(bc+b+1)+1/(ca+c+1)
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提问:若abc=1,求1/(ab+a+1) 1/(bc+b+1)+1/(ca+c+1)
提问:若abc=1,求1/(ab+a+1) 1/(bc+b+1)+1/(ca+c+1)
提问:若abc=1,求1/(ab+a+1) 1/(bc+b+1)+1/(ca+c+1)
abc=1
1/(ab+a+1)+1/(bc+b+1)+1/(ca+c+1)
=1/(1/c+a+1)+1/(1/a+b+1)+1/(1/b+c+1)
=c/(1+ac+c)+a/(1+ab+a)+b/(1+bc+b)
=abc/(ab+a+1)+a/(ab+a+1)+ab/(ab+a+1)
=(abc+ab+a)/(ab+a+1)
=(ab+a+1)/(ab+a+1)
=1
楼上的结论是对的,只不过他没推导出
1/(ab+a+1)+1/(bc+b+1)+1/(ca+c+1)=c/(1+ac+c)+a/(1+ab+a)+b/(1+bc+b)
若abc=1 求a/(ab+a+1) + b/(bc+b+1 )+ c/(ca+c+1) ? ? ?的值
原式=a/(ab+a+1)+ab/(abc+ab+a)+c/(ca+c+abc)
=a/(ab+a+1)+ab/(1+ab+a)+c/[c(a+1+ab)]
=a/(ab+a+1)+ab/(ab+a+1)+1/(ab+a+1)
=(ab+a+1)/(ab...
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若abc=1 求a/(ab+a+1) + b/(bc+b+1 )+ c/(ca+c+1) ? ? ?的值
原式=a/(ab+a+1)+ab/(abc+ab+a)+c/(ca+c+abc)
=a/(ab+a+1)+ab/(1+ab+a)+c/[c(a+1+ab)]
=a/(ab+a+1)+ab/(ab+a+1)+1/(ab+a+1)
=(ab+a+1)/(ab+a+1)
=1
收起
原式=1/(ab+a+1)+a/(abc+ab+a)+1/(ca+c+abc)
=1/(ab+a+1)+a/(1+ab+a)+abc/(ca+c+abc)
=1/(ab+a+1)+a/(ab+a+1)+ab/(a+1+ab)
=(1+a+ab)/(ab+a+1)
=1
清楚了吗?呵呵。。。