(2x ^2+ax-y+b)-(2bx^2-3x+5y-1)的值与x取值无关,求(3a^2-ab-b^2)-(4a^2+ab+b^2)的值
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(2x ^2+ax-y+b)-(2bx^2-3x+5y-1)的值与x取值无关,求(3a^2-ab-b^2)-(4a^2+ab+b^2)的值
(2x ^2+ax-y+b)-(2bx^2-3x+5y-1)的值与x取值无关,求(3a^2-ab-b^2)-(4a^2+ab+b^2)的值
(2x ^2+ax-y+b)-(2bx^2-3x+5y-1)的值与x取值无关,求(3a^2-ab-b^2)-(4a^2+ab+b^2)的值
(2x ^2+ax-y+b)-(2bx^2-3x+5y-1)
=2x ^2+ax-y+b-2bx^2+3x-5y+1
=(2-2b)x ^2+(a+3)x-6y+b+1
和x无关则x系数为0
所以2-2b=0,a+3=0
a=-3,b=1
所以原式=3a²-ab-b²-4a²-ab-b²
=-a²-2ab-2b²
=-9+6-2
=-5
(2x²+ax-y+b)-(2bx²-3x+5y-1)
=2x²+ax-y+b-2bx²+3x-5y-1
=(2-2b)x²+(a+3)x-4y-b-1
∵多项式的值与x的取值无关
∴2-2b=0,a+3=0
∴a=-3,b=1
∴(3a²-ab-b²)-(4a²+ab+...
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(2x²+ax-y+b)-(2bx²-3x+5y-1)
=2x²+ax-y+b-2bx²+3x-5y-1
=(2-2b)x²+(a+3)x-4y-b-1
∵多项式的值与x的取值无关
∴2-2b=0,a+3=0
∴a=-3,b=1
∴(3a²-ab-b²)-(4a²+ab+b²)
=3a²-ab-b²-4a²-ab-b²
=-a²-2ab-2b²
=-(-3)²-2×(-3)×1-2×1²
=-9+6-2
=-5
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