急急谁会初一数学
已知:y3-x3=5,x2y+xy2=6, 求式子3(x2y-5/3xy2)-2(-x3-3-y3)+2(5-2x3)-(7x2y-xy2)的值
已知:y3-x3=5,x2y+xy2=6, 3(x2y-5/3xy2)-2(-x3-3-y3)+2(5-2x3)-(7x2y-xy2) =-2x^3+2y^3-4x^2y-4xy^2+16 =2(y^3-x^3)-4(x^2y+xy^2)+16 =10-24+16 =2
原式=3(x2y)-5(xy2)+2(x3)+6+2(y3)+10-4(x3)-7(x2y)+(xy2) =2(y3)-2(x3)-4(x2y)-4(xy2)+16 =10-24+16 =2
答:1、如果关于字母x 的二次多形式-3x2 +mx+nx2-x+3的值与x的取值无关,求(m+n)(m-n)的值。 合并成 (n-3)x² + (m-1...详情>>
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