计算 (a-b)/(2a+2b)·(a^2+b^2)/(a^2-b^2)
问题描述：
计算 (a-b)/(2a+2b)·(a^2+b^2)/(a^2-b^2)
还有一题[(x+1)^2(1-x)^2]/[(x^2-1)^2]÷[(x-1)^2]/x^2-1
答
(a-b)/(2a+2b)·(a^2+b^2)/(a^2-b^2)
=(a-b)/2(a+b)×(a²+b²)/(a+b)(a-b)
=(a²+b²)/2(a+b)²;
[(x+1)^2(1-x)^2]/[(x^2-1)^2]÷[(x-1)^2]/x^2-1
=（x+1）²×(x+1)(x-1)/(x-1)²;
=(x+1)³/(x-1);
手机提问的朋友在客户端右上角评价点【满意】即可.第二题能详细点吗[(x+1)^2(1-x)^2]/[(x^2-1)^2]÷[(x-1)^2]/x^2-1
=（x²-1）²/(x²-1)²×(x²-1)/(x-1)²
=（x-1)(x+1)/(x-1)²;
=(x+1)/(x-1);
抱歉，上面刚写错了[(x+1)^2(1-x)^2为什么会变成（x²-1）²（x+1)^2(1-x)^2
=(x+1)²(x-1)²
=[(x+1)(x-1)]²
=(x²-1)²;