(a-b)/[a-2√(ab)+b]÷[a+2√(ab)+b]/1
=(√a-√b)(√a+√b)/(√a-√b)² ×1/(√a+√b)²
=1/[(√a-√b)(√a+√b)]
=1/(a-b)(√a-√b)² 和(√a+√b)²你确定可以约=(√a-√b)(√a+√b)/(√a-√b)² ×1/(√a+√b)²分子有(√a-√b)，分母有(√a-√b)²约掉1次还剩1/(√a-√b)分子有(√a+√b)，分母有(√a+√b)²约掉1次还剩1/(√a+√b)∴约完后1/[(√a-√b)(√a+√b)]