(2a+2b分之a-b)-(a²-b²分之a²+b²)
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![(2a+2b分之a-b)-(a²-b²分之a²+b²)](/uploads/image/z/1268172-36-2.jpg?t=%282a%2B2b%E5%88%86%E4%B9%8Ba-b%29-%28a%26%23178%3B-b%26%23178%3B%E5%88%86%E4%B9%8Ba%26%23178%3B%2Bb%26%23178%3B%29)
(2a+2b分之a-b)-(a²-b²分之a²+b²)
(2a+2b分之a-b)-(a²-b²分之a²+b²)
(2a+2b分之a-b)-(a²-b²分之a²+b²)
解
原式
=[(a-b)/2(a+b)]-[(a²+b²)/(a-b)(a+b)]
=(a-b)²/2(a+b)(a-b)-2(a²+b²)/(a-b)(a+b)
=(a²-2ab+b²-2a²-2b²)/(a-b)(a+b)
=(-a²-2ab-b²)/(a-b)(a+b)
=-(a+b)²/(a-b)(a+b)
=-(a+b)/(a-b)