Question

化简求值(8x^2-2y^2)÷(4x^*y+4xy^2+y^3/2y^2-5xy-3x^2) ×(y/y^2-5xy+6x^2)x=-1/2y=1/4

快啊

Answer 1

(8x^2-2y^2)÷[(4x^2*y+4xy^2+y^3)/(2y^2-5xy-3x^2)] ×y/(y^2-5xy+6x^2)
=2(4x^2-y^2)÷[y(4x^2+4xy+y^2)/(2y^2-5xy-3x^2)] ×y/(y-2x)(y-3x)
=2(4x^2-y^2)÷[y(2x+y)^2/(2y+x)(y-3x)] ×y/(y-2x)(y-3x)
=2(4x^2-y^2)×(2y+x)(y-3x)/y(2x+y)^2 ×y/(y-2x)(y-3x)
=2(4x^2-y^2)/(2x+y) ×1/(y-2x)
=2(2x-y)(2x+y)/(2x+y) (y-2x)
=2(2x-y)/(y-2x)
=-2(2x-y)/(2x-y)
=-2

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