Question

若代数式(2x^2+ax-y+6)-(2bx^2+3x+5y+1)的值与字母x的值无关,求代数式5ab^2-[2a^2b-(2a^2b-ab^2)]的值

求步骤求解!!!

Answer 1

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-1
∵代数式的值与x无关
∴2-2b=0且a-3=0
解得a=3且b=1
5ab^2-[2a^2b-(2a^2b-ab^2)]
=5ab2-(2a^2b-2a^2b ab2)
=4ab^2
当a=3，b=1时
原式=4*3*1^2
=12

Answer 2

(2x^2+ax-y+6)-(2bx^2+3x+5y+1)
=2x^2+ax-y+6-2bx^2-3x-5y-1
=(2x^2-2bx^2)+(ax-3x)-(y+5y)+(6-1)
=（2-2b)x^2+(a-3)x-6y+5
∵与x的取值无关
∴
2-2b=0
a-3=0
解得：
b=1
a=3
5ab^2-[2a^2b-(2a^2b-ab^2)]
=5ab^2-[2a^2b-2a^2b+ab^2]
=5ab^2-ab^2
=4ab^2
=4×3×1^2
=12

Answer 3

(2x^2+ax-y+6)-(2bx^2+3x+5y+1)
与x无关，即2x^2+ax-2bx^2-3x=0 （2-2b）x^2+（a-3）x^2=0 2-2b=0 a-3=0 b=1 a=3
代入 5ab^2-[2a^2b-(2a^2b-ab^2)]=4ab^2=12