[3x^2-2x(x+y)+y(2x-y)]÷(x+y),其中X=1/2,y=1/3 先化简,在求值
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[3x^2-2x(x+y)+y(2x-y)]÷(x+y),其中X=1/2,y=1/3 先化简,在求值
[3x^2-2x(x+y)+y(2x-y)]÷(x+y),其中X=1/2,y=1/3 先化简,在求值
[3x^2-2x(x+y)+y(2x-y)]÷(x+y),其中X=1/2,y=1/3 先化简,在求值
[3x^2-2x(x+y)+y(2x-y)]÷(x+y)
=[3x^2-2x^2-2xy+2xy-y^2]÷(x+y)
=(x^2-y^2)÷(x+y)
=(x+y)(x-y)÷(x+y)
=x-y
=1/2-1/3
=1/6
[3x^2-2x(x+y)+y(2x-y)]÷(x+y)
=[x(3x-2x-2y)+y(2x-y)]÷(x+y)
=[x(x-2y)+y(2x-y)]÷(x+y)
=[x^2-2xy+2xy-y^2)]÷(x+y)
=[]x^2-y^2]÷(x+y)
=(x+y)(x-y)÷(x+y)
=(x-y)
=1/2-1/3
=1/6
原式=(3x^2-2x^2-2xy+2xy-y^2)/(x+y)=(x^2-y^2))/(x+y)=x-y
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