已知A/x(x-1)(x+2)=B/x + 1/(x-1) + C/(x+2),求A,B和C的值!
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已知A/x(x-1)(x+2)=B/x + 1/(x-1) + C/(x+2),求A,B和C的值!
已知A/x(x-1)(x+2)=B/x + 1/(x-1) + C/(x+2),求A,B和C的值!
已知A/x(x-1)(x+2)=B/x + 1/(x-1) + C/(x+2),求A,B和C的值!
把方程右边通分,得[B(x-1)(x+2)+x(x+2)+Cx(x-1)]/x(x-1)(x+2)=[(B+1+C)x^2+(B+2-C)x-2B]/x(x-1)(x+2),所以B+1+C=0,B+2-C=0,-2B=A,解得A=3,B=-3/2,C=1/2
A=3 B=-1 C=-1
等式右边通分,分子为(B+C+1)X^2+(B+2-C)X-2B=左边的分子A
B+C+1=0
B+2-C=0
-2B=A
解得
A=3
B=-1·5
C=0·5
通分化简后得
A=B(X-1)(X+2)+X2+2X+CX2-C
A=(B+C+1)X2+(B+2)X+(-2B-C)
方程组:B+C+1=0
B+2=0
-2B-C=A
解得A=3,B=-2,C=1
注:X2表示X的平方,下面的以此类推
不懂的可以再问,我在线
A/x(x-1)(x+2)=B/x + 1/(x-1) + C/(x+2)
A/x(x-1)(x+2)=[B(x-1)(x+2)+x(x+2)+Cx(x-1)]/x(x-1)(x+2)
=[B(x^2+x-2)+x^2+2x+Cx^2-Cx]/x(x-1)(x+2)
A=Bx^2+Bx-2B+x^2+2x+Cx^2-Cx
=(B+1+C)x^2+(B+2-C)x-2B
B+1+C=0
B+2-C=0
A=-2B
C=1/2
B=-3/2
A=3
对右边变形得:
[B(x-1)(x+2)+x(x+2)+Cx(x-1)]/x(x-1)(x+2)
这样分母和左边一样,下面对上式的分子进行变形整理得:(B+C+1)x^2+(B-C+2)x-2B.
由原式知:(B+C+1)x^2+(B-C+2)x-2B=A.
所以:B+C+1=0
B-C+2=0
-2B=A
所以:A=3,B=-3/2,C=1/2.
:A/[x(x-1)(x+2)]=[B(x-1)(x+2)]/[x(x-1)(x+2)]+[x(x+2)]/[x(x-1)(x+2)]+[Cx(x-1)]/[x(x-1)(x+2)]
A/[x(x-1)(x+2)]=[(B+C+1)x²+(B-C+2)x-2]/x(x-1)(x+2)
∴A=-2B,B+C+1=0,B-C+2=0, 解得B=-1.5,C=0.5
∴A=3,B=-1.5,C=0.5