数学已知3X^2+2x+4=a(x-1)(x+2)+b(x-1)-c,求a,b,c的值
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数学已知3X^2+2x+4=a(x-1)(x+2)+b(x-1)-c,求a,b,c的值
数学已知3X^2+2x+4=a(x-1)(x+2)+b(x-1)-c,求a,b,c的值
数学已知3X^2+2x+4=a(x-1)(x+2)+b(x-1)-c,求a,b,c的值
右边=ax²+ax-2a+bx-b-c
=ax²+(a+b)x-2a-b-c
=3x²+2x+4
x次数相同的项系数相等
所以a=3
a+b=2
-2a-b-c=4
所以
a=3
b=-1
c=-9
a(x-1)(x+2)+b(x-1)-c=ax^2+ax-2a+bx-b-c
=ax^2+(a+b)x-2a-b-c=3x^2+2x+4
∴a=3,a+b=2,-2a-b-c=4
a=3,b=-1,c=-9
3X^2+2x+4=a(x-1)(x+2)+b(x-1)-c
=a(x^2 +x -2)+bx-b-c
=ax^2 +(a+b)x -(2a+b+c)
所以a=3 b=-1 c=-9
这是待定系数法,是正统方法
a(x-1)(x+2)+b(x-1)-c=ax^2+ax-2a+bx-b-c=ax^2+(a+b)x-2a-b-c=3x^2+2x+4
∴a=3,a+b=2,-2a-b-c=4
a=3,b=-1,c=-9
3x平方+2x+4=a(x-1)(x+2)+b(x-1)-c=a(x平方+x-2)+bx-b-c=ax平方+ax-2a+bx-b-c=ax平方+(a+b)x-(2a+b+c)所以有a=3,a+b=2,-(2a+b+c)=4解得a=3,b=-1,c=-9
本题可用待定系数法来求解
3X^2+2x+4=a(x-1)(x+2)+b(x-1)-c
=a(x^2 +x -2)+bx-b-c
=ax^2 +(a+b)x -(2a+b+c)
由观察可知 a=3 b=-1 c=-9
这也上来求解啊,强悍啊