(1-(a+1分之2))的平方除以(a+1)分之(a-1)

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(1-(a+1分之2))的平方除以(a+1)分之(a-1)

(1-(a+1分之2))的平方除以(a+1)分之(a-1)
(1-(a+1分之2))的平方除以(a+1)分之(a-1)

(1-(a+1分之2))的平方除以(a+1)分之(a-1)
原式=[(a+1-2)/(a+1)]²×[(a+1)/(a-1)]
=[(a-1)/(a+1)]²×[(a+1)/(a-1)]
=[(a-1)²/(a+1)²]×[(a+1)/(a-1)]
=(a-1)/(a+1)

[1-2/(a+1)]^2/[(a-1)/(a+1)]
=[(a+1-2)/(a+1)]^2/[(a-1)/(a+1)]
=(a-1)^2/(a+1)^2*(a+1)/(a-1)
=(a-1)/(a+1)

[1-2/(a+1)]^2/[(a-1)/(a+1)]
=[(a-1)/(a+1)]^2/[(a-1)/(a+1)]
=[(a-1)/(a+1)]^2*(a+1)/(a-1)
=(a-1)/(a+1)

上面的式子通分,然后化简就好,不明白再联系