2條數!急!(要步驟)

1.
a + 1/(a+1) = b + 1/(b-1) - 2
0 = b - a - 2 + 1/(b-1) - 1/(a+1) 【全部移到右邊】
0 = -(a-b+2) + 1/(b-1) - 1/(a+1) 【抽出 -1】
0 = -(a-b+2) + [(a+1) - (b-1)]/(b-1)(a+1) 【通分母】
0 = -(a-b+2) + (a-b+2)/(b-1)(a+1)
0 = -1 + 1/(b-1)(a+1) 【全式除以(a-b+2)，因題目給定a-b+2≠0】
1/(b-1)(a+1) = 1
(b-1)(a+1) = 1
ab + b - a - 1 = 1 【展開】
ab - a + b = 2


2.
相信你應該打錯了題目，由題目的循環性，題目應為：
a/(ab+a+1) + b/(bc+b+1) + c/(ca+c+1)
= a/(ab+a+1) * c/c + b/(bc+b+1) * a/a + c/(ca+c+1) * b/b 【每項均乘以一個等於1的數字】
= ac/(abc+ac+c) + ba/(abc+ab+a) + bc/(abc+bc+b)
= ac/(ac+c+1) + ba/(ba+a+1) + bc/(bc+b+1) ----------------------(1) 【因為abc=1】
= ac/(ac+c+1) * b/b + ba/(ba+a+1) *c/c + bc/(bc+b+1) *a/a 【每項均乘以一個等於1的數字】
= abc/(abc+bc+b) + abc/(abc+ac+c) + abc/(abc+ab+a)
= 1/(bc+b+1) + 1/(ac+c+1) + 1/(ab+a+1) ----------------------(2) 【因為abc=1】

從(1)和(2)，我們知道
a/(ab+a+1) + b/(bc+b+1) + c/(ca+c+1)
= ac/(ac+c+1) + ba/(ba+a+1) + bc/(bc+b+1)
= 1/(bc+b+1) + 1/(ac+c+1) + 1/(ab+a+1)

因此，
a/(ab+a+1) + b/(bc+b+1) + c/(ca+c+1)
= 1/3*{[a/(ab+a+1) + b/(bc+b+1) + c/(ca+c+1)] + [a/(ab+a+1) + b/(bc+b+1) + c/(ca+c+1)] + [a/(ab+a+1) + b/(bc+b+1) + c/(ca+c+1)]} 【將該項複制3份後再乘以1/3】
= 1/3*{[a/(ab+a+1) + b/(bc+b+1) + c/(ca+c+1)] + [ac/(ac+c+1) + ba/(ba+a+1) + bc/(bc+b+1)] + [1/(bc+b+1) + 1/(ac+c+1) + 1/(ab+a+1)]} 【將其中兩項分別換為(1)和(2)】
= 1/3*{[a/(ab+a+1) + ba/(ba+a+1) + 1/(ab+a+1)] + [b/(bc+b+1) + bc/(bc+b+1) + 1/(bc+b+1)] + [c/(ca+c+1) + ac/(ac+c+1) + 1/(ac+c+1)]} 【將各項重新排列】
= 1/3*[(a+ab+1)/(ab+a+1) + (b+bc+1)/(bc+b+1) + (c+ac+1)/(ac+c+1)]
= 1/3*(1+1+1)
= 1

希望幫倒你！ ^^