3{x}^{2}+kxy-8{y}^{2}除以x-2y,商为3x+4y则k的值为?
问题描述:
3{x}^{2}+kxy-8{y}^{2}除以x-2y,商为3x+4y则k的值为?
答
(x-2y)(3x+4y)
=3x^2-2xy-8y^2
3x^2+kxy-8y^2
k=-2
答
∵(x-2y)(3x+4y)=3x²-2xy-8y²
∴3x²+kxy-8y²=3x²-2xy-8y²
∴k=-2
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