两道数学题,初二高手来呀!如下图 如果回答的好 我加双倍的分

[36a^(4m+5)-18a^(2m+3)*a^(m-1)]÷[(-3a^m)^2*(-2a^2)]
=[36a^(4m+5)-18a^(2m+3+m-1)]÷[(-9/2)*a^(2m+2)]
=[36a^(4m+5)-18a^(3m+2)]÷[(-9/2)*a^(2m+2)]
=[36a^(4m+5)]/[(-9/2)*a^(2m+2)]-18a^(3m+2)/[(-9/2)*a^(2m+2)]
=-8a^(2m+3) + 4a^m
[0.25a^2b-(1/2)a^3b^2-(1/6)a^4b^3]÷(-0.5a^2b)
=[0.25a^2b/(-0.5a^2b)]-[(1/2)a^3b^2/(-0.5a^2b)]-[(1/6)a^4b^3/(-0.5a^2b)]
=-0.5+ab+(1/3)a^2*b^2

1:原式=[36a^(4m+5)-18^(3m+2）]÷[-18a^(2m+2）]
=-2a^(2m+3)+a^(m)
2:原式=-2/1+ab+1/3a^(2)b^(2)

看不见

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