已知x(x-1)(x+2)分之x的平方+2=x分之A+x-1分之B+x+2分之C,求A、B、C的值.
来源:学生作业帮助网 编辑:作业帮 时间:2024/11/02 21:38:11
已知x(x-1)(x+2)分之x的平方+2=x分之A+x-1分之B+x+2分之C,求A、B、C的值.
已知x(x-1)(x+2)分之x的平方+2=x分之A+x-1分之B+x+2分之C,求A、B、C的值.
已知x(x-1)(x+2)分之x的平方+2=x分之A+x-1分之B+x+2分之C,求A、B、C的值.
(x^2+2)/x(x-1)(x+2)=A/x+B/(x-1)+C/(x+2)
= {(A+B+C)X^2+(A+2B-C)X-2A} /x(x-1)(x+2)
根据系数相等:
A+B+C=1
A+2B-C=0
-2A=2
解上述联立方程,得 A= -1,B=1,C=1
A/x +B/(x-1)+C/(x+2)
=[A(x-1)(x+2)+Bx(x+2)+Cx(x-1)]/[x(x-1)(x+2)]
=[A(x²+x-2)+B(x²+2x)+C(x²-x)]/[x(x-1)(x+2)]
=[(A+B+C)x²+(A+2B-C)x-2A]/[x(x-1)(x+2)]
=(x²+2)/[...
全部展开
A/x +B/(x-1)+C/(x+2)
=[A(x-1)(x+2)+Bx(x+2)+Cx(x-1)]/[x(x-1)(x+2)]
=[A(x²+x-2)+B(x²+2x)+C(x²-x)]/[x(x-1)(x+2)]
=[(A+B+C)x²+(A+2B-C)x-2A]/[x(x-1)(x+2)]
=(x²+2)/[x(x-1)(x+2)]
A+B+C=1 (1)
A+2B-C=0 (2)
-2A=2 (3)
(1)+(2)
2A+3B=1
B=(1-2A)/3
由(3)得A=-1
B=[1-2(-1)]/3=1
A=-1 B=1代入(1)
C=1
A=-1 B=1 C=1
收起