Question

已知关于x,y的多项式ax的平方+2bxy+x的平方-x-2xy+y不含二次项,求5a-8b的值。

Answer 1

ax^2+2bxy+x^2-x-2xy+y
=(a+1)x^2+(2b-2)xy-x+y
因为不含二次项
所以a+1=0 且 2b-2=0
所以a=-1，b=1
所以5a-8b=-5-8=-13
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Answer 2

由ax²+2bxy+x²-x-2xy+y
=（ax²+2bxy+x²+2xy）-x+y
不含二次项，
∴a+1=0（1）得a=-1
2b+2=0（2）得b=-1
∴5a-8b=3.

Answer 3

ax^2+2bxy+x^2-x-2xy+y
ax^2+2bxy+x^2-2xy=(a+1)x^2 + 2(b-1)xy = 0
a=-1 b=1
5a-8b=5×(-1) - 8×(1) = -5 - 8 = - 13

Answer 4

ax^2+2bxy+x^2-x-2xy+y
=(a+1)x^2+(2b-2)xy-x+y
因为不含二次项
所以a+1=0 且 2b-2=0
所以a=-1，b=1
所以5a-8b=-5-8=-13
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Answer 5

(a+1)x^2+(2b-2)xy-x+y
a+1=0 2b-2=0
a=-1 b=1
5a-8b=-5+8=3

Answer 6

因为(ax)^2+2bxy+x^2-x-2xy+y不含二次项
则a+1=0 & 2b-2=0
a=-1 & b=1
5a-8b=5*(-1)-8*(1)=-13