lm+n-3l+m^2n^2+4mn+4=0
lm+n-3l+(mn+2)²=0
則：
m+n-3=0
mn+2=0
解得：
m+n=3
mn=-2
m²+n²=(m+n)²-2mn=9+4=13
(4m-n)(m-2n)-(n-m）(-m-n)
=4m²-9mn+2n²+n²-m²
=3m²+3n²+18
=3(m²+n²)+18
=39+18
=57
lm+n-3l+m^2n^2+4mn+4=0 求(4m-n)(m-2n)-(n-m）(-m-n)
lm+n-3l+m^2n^2+4mn+4=0 求(4m-n)(m-2n)-(n-m）(-m-n)
數(shù)學(xué)人氣：127 ℃時間：2020-08-15 02:35:45
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