初二数学题，聪明的看下~
\u51e0\u9053\u521d\u4e8c\u6570\u5b66\u9898,\u806a\u660e\u7684\u8bf7\u6765!(1)C.44\u4e2a
(2)\u5927\u4e8e
\uff083\uff09A\u5c0f\u4e8eB
\uff084\uff093\uff0c-6\uff0c\u6839\u53f72\uff0c-\u6839\u53f78 \u6839\u53f72*-\u6839\u53f78+3-\uff08-6\uff09
\u76ee\u524d\u662f\u5168\u90e8\u6b63\u786e\u7684\u3002
括号里通分，{[（a-2）(a+2)]/[a(a+2)^2]-[(a-1)a]/[a(a+2)^2]}*(a-4)/(a+2)=(a^2-4-a^2+a)/[a(a+2)^2]*(a-4)/(a+2)=1/(a^2+2)
因为a^2+2a-1=0,所以a^2+2a=1，所以上式等于1
原式=[(a-2)/(a+2)*a-(a-1)/(a+2)^2]*(a+2)/(a-4)
=[(a-2)/a-(a-1)/(a+2)]*1/(a-4)
=[(a-2)(a+2)-(a-1)*a]/(a-4)*(a+2)*a
=[a^2-4-a^2+a]/(a-4)*(a+2)*a
=[a-4]/(a-4)*(a+2)*a
=1/a(a+2)
=1
=((a-2)(a+2)-a(a-1))/a(a+2)^2/((a-4)/(a+2))
=(a-4)/a(a+2)^2/(a-4)*(a+2)
=1/a(a+2)
=1/(a^2+2a)
=1/1=1
={（a-2)/a(a+2)-(a-1)/(a+2)~2} / {(a-4)/(a+2)}
={(a+2)(a-2) / a(a+2)~2 - a(a-1) / a(a+2)~2} / {(a-4)/(a+2)}
={(a~2-4-a~2+a) / a(a+2)~2 }/ {(a-4)/(a+2)}
=(a-4) / a(a+2)~2 * (a+2)/(a-4)
=1/ {a(a+2)}
=1
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