化简:(1)(a+2b)/(a+b)+2b的平方/a的平方-b的平方 (2)(x的平方-1)/(x的平方+2x+1)-(x+1)/(x-1)初

问题描述:

化简:(1)(a+2b)/(a+b)+2b的平方/a的平方-b的平方 (2)(x的平方-1)/(x的平方+2x+1)-(x+1)/(x-1)初

解1题
原式=(a+2b)/(a+b) +2b²/(a²-b²) 通分
=(a+2b)(a-b)/[(a+b)(a-b)] +2b²/[(a+b)(a-b)]
=[(a+2b)(a-b)+2b²]/[(a+b)(a-b)]
=(a²-ab+2ab-2b²+2b²)/[(a+b)(a-b)]
=(a²+ab)/[(a+b)(a-b)]
=a(a+b)/[(a+b)(a-b)] 约分
=a/(a-b)
解2题
原式=(x²-1)/(x²+2x+1) -(x+1)/(x-1) 分解因式
=(x+1)(x-1)/(x+1)² -(x+1)/(x-1) 约分
=(x-1)/(x+1) -(x+1)/(x-1) 通分
=(x-1)²/[(x+1)(x-1)] -(x+1)²/[(x+1)(x-1)]
=[(x-1)²-(x+1)²]/[(x+1)(x-1)]
=(x²-2x+1-x²-2x-1)/[(x+1)(x-1)]
=-4x/[(x+1)(x-1)]