已知(3x+1)(3x-3)-3x+2)^2=-7,(y+7)(y-7)-y(y-7)=0,求8y^2-5y(-y+3x
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已知(3x+1)(3x-3)-3x+2)^2=-7,(y+7)(y-7)-y(y-7)=0,求8y^2-5y(-y+3x)+4y(-4y-5/2x)的值
(3x+1)(3x-3)-(3x+2)^2
=9x^2-6x-3-(9x^2+12x+4)
=-18x-7=-7
∴ x=0
(y+7)(y-7)-y(y-7)
=y^2-49-y^2+7y
=7y-49=0
∴ y=7
8y^2-5y(-y+3x)+4y(-4y-5/2x)
=8y^2+5y^2-15xy-16y^2-10xy
=-3y^2-25xy
=-3*7*7-0
=-147
=9x^2-6x-3-(9x^2+12x+4)
=-18x-7=-7
∴ x=0
(y+7)(y-7)-y(y-7)
=y^2-49-y^2+7y
=7y-49=0
∴ y=7
8y^2-5y(-y+3x)+4y(-4y-5/2x)
=8y^2+5y^2-15xy-16y^2-10xy
=-3y^2-25xy
=-3*7*7-0
=-147
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