Question

已知（ax+b)(x+2)=2(x+1)^2-(x+4),求a+2b的值

是（X+1)^2 谢谢

Answer 1

已知（ax+b)(x+2)=2(x+1)^2-(x+4),
则ax²+(2a+b)x+2b=2x²+4x+2-x-4=2x²+3x-2
两边各项系数相等
则a=2
2a+b=3
2b=-2
解得a=2 b=-1

Answer 2

展开 ax^2 + (2a+b)x +2b = 2x^2 +3x -2
a =2 b= -1
a + 2b = 0

Answer 3

应该是a=2，b=一1