2026-4-18數學動動腦

已知

1/(a+1)+1/(b+1)+1/(c+1)=1

求abc=?

-->如下參考:

(b+1)(c+1)+(a+1) (c+1)+(a+1)(b+1) / (a+1) (b+1)(c+1) = 1

(b+1)(c+1)+(a+1) (c+1)+(a+1)(b+1)=(a+1) (b+1)(c+1) 

L

(b+1)(c+1)+(a+1) (c+1)+(a+1)(b+1)

                      = (bc+b+c+1)+(ac+a+c+1)+(ab+a+b+1)  

                      =bc+ac+ab+b+c+a+c+a+b+1+1+1

                      =bc+ac+ab+3+3+3

R

(a+1) (b+1)(c+1)

                        =(ab+a+b+1)(c+1)

                        =abc+ab+ac+a+bc+b+c+1

                       =abc+ab+ac+bc+3+1

L=R

兩邊消除 bc+ac+ab ，得 9=abc+4

abc=5

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