rachel_qi
幼苗
共回答了15个问题采纳率:93.3% 举报
(2x^2+ax-y+6)-(2bx^2-3x+5y-1)的值与字母x所取得值无关
说明x的系数为0
(2x^2+ax-y+6)-(2bx^2-3x+5y-1)
=(2-2b)x^2+(a+3)x-(1+5)y+7
所以x的系数为0
2-2b=0,a+3=0
=>a=-3,b=1
1/3*a^3-2b^2-(1/4*a3-3b)
=1/3*a^3-2b^2-1/4*a63+3b
=1/12*a^3-2b^2+3b
=1/12*27-2*1+3*1
=39/12
1年前
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