这一道初二数学题，我非常d不明白。（拜托写下过程·-·）

(m-3/2m-4)÷(m+2-5/m-2)
解:原式=(m-3/2m-4)(m-2/m-3) "乘倒数."
=(1/2m-4)(m-2)"交叉约分"
=[1/2(m-2)](m-2)"提分母中的2"
=1/2 "m-2约掉了."

(m-3)/(2m-4 )÷ [m+2-5/(m-2）]
=(m-3)/(2m-4 )÷ (m^2-9)/(m-2)
=[(m-3)/2(m-2 )] [(m-2)/(m+3)(m-3)]
=1/(2m+6)

先算除号后面括号里的部分，先通分得：（m+2）(m-2)/(m-2)-5/(m-2)=(m平方-4)/(m-2)-5/(m-2)=(m平方-9)/(m-2),而m平方-9=(m-3)(m+3).所以
（m-3）/（2m-4 ）÷ （m+2-5/m-2）=（m-3）/2（m-2）÷ ((m-3)(m+3)/(m-2)),所以等于1/2(m+3).

1/2
解：原式=(m-3/2m-4)×(m-2/m-3)
=1/2