因为 代数式(2x的2次方+ax-y+6)-(2bx的2次方-3x+5y-1)的值与字母x的取值无关
则 x^2, x的系数均为零
x^2的系数 2-2b=0 b=1
x的系数 a+3=0 a=-3
1/3a3次方-2b的2次方
=1/3(-3)^3-2*1^2
=-13
-(1/4a3次方-2b的2次方)
=-(1/4)(-3)^3+2*1^2
=43/4
欢迎追问!
解:(2x^2+ax-y+6)-(2bx^2-3x+5y-1)
=2x^2+ax-y+6-2bx^2+3x-5y+1
=(2-2b)x^2+(a+3)x-6y+7
∴2-2b=0,b=1
∵a+3=0,a=-3
∴3(a^2-2ab-b^2)-//
3/2(2a^2-5ab+2b^2)=3a^2-6ab-3b^2-3a^2+15/2ab-3b^2=3/2ab-6b^2=-9/2-6=-/
21/2.
1/3a3次方-2b的2次方
=1/3(-3)^3-2*1^2
=-13
-(1/4a3次方-2b的2次方)
=-(1/4)(-3)^3+2*1^2
=4-13+43/4
=-9/43/4