![](https://iknow-base.cdn.bcebos.com/lxb/notice.png)
HELP ME!
七.综合运用题(2问4分)28.已知,(2X-1)2=AX2+BX+C求A+B+C的值。解:∵当x=1时,AX2+BX+C=A+B+C∴A+B+C=(2×1-1)2=1(...
七.综合运用题(2问4分)
28.已知 ,(2X-1)2=AX2+BX+C求A+B+C的值。
解:∵当x=1时,AX2+BX+C=A+B+C
∴A+B+C=(2×1-1)2=1
(1) 用类似的方法,求出A-B+C的值;
(2) 试求出下列式子的值:A+C _____________
4A+B+4C= _____________ 展开
28.已知 ,(2X-1)2=AX2+BX+C求A+B+C的值。
解:∵当x=1时,AX2+BX+C=A+B+C
∴A+B+C=(2×1-1)2=1
(1) 用类似的方法,求出A-B+C的值;
(2) 试求出下列式子的值:A+C _____________
4A+B+4C= _____________ 展开
2个回答
展开全部
(1)
∵当 x = -1 时,AX² + BX + C = A - B + C
∴A - B + C = [ 2×(-1) - 1]² = 9
(2)
∵ A + B + C = 1 , A - B + C = 9
∴ ( A + B + C ) + ( A - B + C ) = 10
∴ 2A + 2C = 10
∴ A + C = 5
∵ A + B + C = 1 , A - B + C = 9
∴ ( A + B + C ) - ( A - B + C ) = -8
∴ 2B = -8
∴ B = -4
∵ A + C = 5
∴4A + B + 4C = 4(A + C) + B = 4 × 5 - 4 = 16
∵当 x = -1 时,AX² + BX + C = A - B + C
∴A - B + C = [ 2×(-1) - 1]² = 9
(2)
∵ A + B + C = 1 , A - B + C = 9
∴ ( A + B + C ) + ( A - B + C ) = 10
∴ 2A + 2C = 10
∴ A + C = 5
∵ A + B + C = 1 , A - B + C = 9
∴ ( A + B + C ) - ( A - B + C ) = -8
∴ 2B = -8
∴ B = -4
∵ A + C = 5
∴4A + B + 4C = 4(A + C) + B = 4 × 5 - 4 = 16
推荐律师服务:
若未解决您的问题,请您详细描述您的问题,通过百度律临进行免费专业咨询