作业帮(1)已知x+y=6,xy=-3,x^2y+xy^2=(2)x*[2-(1/x)]+(x/x^2-2x)*(x^2-4)
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(1)已知x+y=6,xy=-3,x^2y+xy^2=(2)x*[2-(1/x)]+(x/x^2-2x)*(x^2-4)
(1)已知x+y=6,xy=-3,x^2y+xy^2=
(2)x*[2-(1/x)]+(x/x^2-2x)*(x^2-4)
(1)已知x+y=6,xy=-3,x^2y+xy^2=(2)x*[2-(1/x)]+(x/x^2-2x)*(x^2-4)
(1)已知x+y=6, xy=-3 x^2y+xy^2 =xy(x+y) =-3*6 =-18 (2)x*[2-(1/x)]+(x/x^2-2x)*(x^2-4) =x*(2x-1)/x+[x/x(x-2)]*(x+2)(x-2) =(2x-1)+(x+2)(x-2)/(x-2) =2x-1+x+2 =3x+1
已知1/x=1/y=6,则3x-2xy+3y/3xy+x+y=?已知1/x+1/y=6,则3x-2xy+3y/3xy+x+y=?
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已知1/x-1/y=2,求3x+7xy-3y/2x-3xy-2y答案是1/x-1/y=2(y-x)/xy=2 得：x-y=-2xy(3x+7xy-3y)/(2x-3xy-2y)=[3(x-y)+7xy]/[2(x-y)-3xy]=(-6xy+7xy)/(-4xy-3xy)=xy/(-7xy)=-1/7 谁能帮我解释一下=[3(x-y)+7xy]/[2(x-y)-3xy]=(-6xy+7xy)/(-4xy-3xy) 是怎
(1)已知x+y=6,xy=-3,x^2y+xy^2=(2)x*[2-(1/x)]+(x/x^2-2x)*(x^2-4)
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已知x+y=6,xy=-3,则x^2y+xy^2=?
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