数学整式问题,速度解决
若(4m的二次方+amn+bn的二次方)除以(m+3n)=4m-2n,求常数a.b的值。 注意:要写过程。 快,速度,谢谢
(4m²+amn+bn²)/(m+3n)=4m—2n 4m²+amn+bn²=4m²+12mn—2mn—6n² 4m²+amn+bn²=4m²+10mn—6n² 推论出 a=10,b=-6
题目写出来: (4m^2+amn+bn^2)/(m+3n) = 4m-2n 两边同乘(m+3n)得: (4m^2+amn+bn^2) = (4m-2n)(m+3n) 即:(4m^2+amn+bn^2) = 4m^2+10mn-6n^2 比较两边各项系数得: a=10,b=-6
(4m^2+amn+bn^2)/(m+3n)=4m-2n, 则4m^2+amn+bn^2=(m+3n)(4m-2n) =4m^2+10mn-6n^2, ∴a=10,b=-6.
(4m^2+amn+bn^2)/(m+3n)=4m-2n (m+3n)(4m-2n)=4m^2+10mn-6n^2=4m^2+amn+bn^2 所以 a=10,b=-6
直接用(4m-2n)*(m+3n)=4m平方+10mn-6n平方,就得出a=10,b=-6
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