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一道初一下册整式除法题.
m^2+9n^2+6m+6n+10=0,求(m^2-6mn+9n^2)÷(m-3n)-(4m^2-9n^2)÷(2m-3n)的值.
人气:371 ℃ 时间:2020-06-04 08:50:05
解答
(m^2-6mn+9n^2)÷(m-3n)- (4m^2-9n^2)÷(2m-3n)=(m-3n)^2/(m-3n)-(2m+3n)*(2m-3n)/(2m-3n)=m-3n-2m-3n=-m-6nm^2+9n^2+6m+6n+10=(m+3)^2+(3n+1)^2=0因为(m+3)^2>=0,(3n+1)^2>=0所以(m+3)^2=0,(3n+1)^2=0m=-3,...
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