1.将已知三式取倒数并通分:
[(a+b)*c+(b+c)*a+(a+c)*b]/(abc)=3+4+5=12
即(2ab+2bc+2ac)/(abc)=12
推出:abc/(ab+bc+ac)=1/6
2.a + 1/a+1 = b + 1/b-1 - 2
(a+1)+1/(a+1)=(b-1)+1/(b-1)
(a+1)-(b-1)=1/(b-1)-1/(a+1)
(a-b+2)=(a-b+2)/[(a+1)(b-1)]
因为a-b+2不等于0,所以等式两边约分得:
1=1/[(a+1)(b-1)]
(a+1)(b-1)=1
ab-a+b-1=1
ab-a+b=2