Question

若代数式(2x^2+ax-y+6)-(2bx^2-3x+5y-1)的值与字母x的取值无关，你能否求出代数

若代数式(2x^2+ax-y+6)-(2bx^2-3x+5y-1)的值与字母x的取值无关，你能否求出代数式0.5a^2-2b^2+4ab的值？

Answer 1

(2x^2+ax-y+6)-(2bx^2-3x+5y-1)=(2-2b)*x^2+(a+3)*x-6y+7
若代数式(2x^2+ax-y+6)-(2bx^2-3x+5y-1)的值与字母x的取值无关
则b=1 a=-3
0.5a^2-2b^2+4ab=4.5-2-12=-9.5

Answer 2

(2x^2+ax-y+6)-(2bx^2-3x+5y-1)
=(2-2b)x²+(a+3)x-6y+7
值与字母x的取值无关
则2-2b=0，a+3=0
则b=1，a=-3
0.5a^2-2b^2+4ab
=0.5×9-2-12
=4.5-2-12
=-9.5

Answer 3

(2x^2+ax-y+6)-(2bx^2-3x+5y-1)=(2-2b)x^2+(a+3)x-6y+7与字母x的取值无关,所以a=-3,b=1
故0.5a^2-2b^2+4ab=0.5x9-2x1+4x1x(-3)=-9.5