1.把分式[x^2-(y-z)^2]/[(x+y)^2-z^2]约分得_________.2.m为何值时,关于x的方程2/(x-2)+mx/(x^2-4)=3/(x-2)会产生增根?3.已知(2x+3)/[x*(x-1)*(x+2)]=A/x+B/(x-1)+C/(x+2) (A.B.C都为常数),求A.B.C的值分别是多少?4.已知a^2-ab-2b^
来源:学生作业帮助网 编辑:作业帮 时间:2024/06/27 13:13:48
![1.把分式[x^2-(y-z)^2]/[(x+y)^2-z^2]约分得_________.2.m为何值时,关于x的方程2/(x-2)+mx/(x^2-4)=3/(x-2)会产生增根?3.已知(2x+3)/[x*(x-1)*(x+2)]=A/x+B/(x-1)+C/(x+2) (A.B.C都为常数),求A.B.C的值分别是多少?4.已知a^2-ab-2b^](/uploads/image/z/8760310-70-0.jpg?t=1.%E6%8A%8A%E5%88%86%E5%BC%8F%5Bx%5E2-%28y-z%29%5E2%5D%2F%5B%28x%2By%29%5E2-z%5E2%5D%E7%BA%A6%E5%88%86%E5%BE%97_________.2.m%E4%B8%BA%E4%BD%95%E5%80%BC%E6%97%B6%2C%E5%85%B3%E4%BA%8Ex%E7%9A%84%E6%96%B9%E7%A8%8B2%2F%28x-2%29%2Bmx%2F%28x%5E2-4%29%3D3%2F%28x-2%29%E4%BC%9A%E4%BA%A7%E7%94%9F%E5%A2%9E%E6%A0%B9%3F3.%E5%B7%B2%E7%9F%A5%282x%2B3%29%2F%5Bx%2A%28x-1%29%2A%28x%2B2%29%5D%3DA%2Fx%2BB%2F%28x-1%29%2BC%2F%28x%2B2%29+%28A.B.C%E9%83%BD%E4%B8%BA%E5%B8%B8%E6%95%B0%29%2C%E6%B1%82A.B.C%E7%9A%84%E5%80%BC%E5%88%86%E5%88%AB%E6%98%AF%E5%A4%9A%E5%B0%91%3F4.%E5%B7%B2%E7%9F%A5a%5E2-ab-2b%5E)
1.把分式[x^2-(y-z)^2]/[(x+y)^2-z^2]约分得_________.2.m为何值时,关于x的方程2/(x-2)+mx/(x^2-4)=3/(x-2)会产生增根?3.已知(2x+3)/[x*(x-1)*(x+2)]=A/x+B/(x-1)+C/(x+2) (A.B.C都为常数),求A.B.C的值分别是多少?4.已知a^2-ab-2b^
1.把分式[x^2-(y-z)^2]/[(x+y)^2-z^2]约分得_________.
2.m为何值时,关于x的方程2/(x-2)+mx/(x^2-4)=3/(x-2)会产生增根?
3.已知(2x+3)/[x*(x-1)*(x+2)]=A/x+B/(x-1)+C/(x+2) (A.B.C都为常数),求A.B.C的值分别是多少?
4.已知a^2-ab-2b^2=0,求a/b-b/a-(a^2+b^2)/ab的值.
1.把分式[x^2-(y-z)^2]/[(x+y)^2-z^2]约分得_________.2.m为何值时,关于x的方程2/(x-2)+mx/(x^2-4)=3/(x-2)会产生增根?3.已知(2x+3)/[x*(x-1)*(x+2)]=A/x+B/(x-1)+C/(x+2) (A.B.C都为常数),求A.B.C的值分别是多少?4.已知a^2-ab-2b^
1.[x^2-(y-z)^2]/[(x+y)^2-z^2]
=(x-y+z)(x+y-z)/(x+y+z)(x+y-z)
=(x-y+z)/(x+y+z)
2.2/(x-2)+mx/(x^2-4)=3/(x-2)
mx/(x^2-4)=1/(x-2)
mx=(x^2-4)/(x-2)
mx=x+2
(m-1)x=2
m=1时,方程会产生增根
3.(2x+3)/[x*(x-1)*(x+2)]=A/x+B/(x-1)+C/(x+2)
2x+3=A(x-1)*(x+2)+Bx*(x+2)+Cx*(x-1)
2x+3=A(x^2+x-2)+B(x^2+2x)+C(x^2-x)
2x+3=(A+B+C)x^2+(A+2B-C)x-2A
A+B+C=0
A+2B-C=2
-2A=3
A=-3/2,B=5/3,C=-1/6
4.a^2-ab-2b^2=0
(a-2b)(a+b)=0
a=2b或a=-b
a=2b
a/b-b/a-(a^2+b^2)/ab
=2-1/2-(4b^2+b^2)/2b^2
=3/2-5/2
=-1
a=-b
a/b-b/a-(a^2+b^2)/ab
=-1+1-2b^2/(-b^2)
=0+2
=2
1.把分式[x^2-(y-z)^2]/[(x+y)^2-z^2]约分得_________.
[x^2-(y-z)^2]/[(x+y)^2-z^2]
=[(x+y-z)(x-y+z)]/[(x+y-z)(x+y+z)]
=(x-y+z)/(x+y+z)
3.已知(2x+3)/[x*(x-1)*(x+2)]=A/x+B/(x-1)+C/(x+2) (A.B.C都为常数...
全部展开
1.把分式[x^2-(y-z)^2]/[(x+y)^2-z^2]约分得_________.
[x^2-(y-z)^2]/[(x+y)^2-z^2]
=[(x+y-z)(x-y+z)]/[(x+y-z)(x+y+z)]
=(x-y+z)/(x+y+z)
3.已知(2x+3)/[x*(x-1)*(x+2)]=A/x+B/(x-1)+C/(x+2) (A.B.C都为常数),求A.B.C的值分别是多少?
(2x+3)/[x*(x-1)*(x+2)]=A/x+B/(x-1)+C/(x+2)
(2x+3)/[x*(x-1)*(x+2)]=[A*(x-1)*(x+2)+B*(x+2)*x+C*(x-1)*x]/[x*(x-1)*(x+2)]
2x+3=A*(x-1)*(x+2)+B*(x+2)*x+C*(x-1)*x
继续解就行了
4.已知a^2-ab-2b^2=0,求a/b-b/a-(a^2+b^2)/ab的值.
a^2-ab-2b^2=0
a^2-b^2-b^2-ab=0
(a+b)*(a-b)-b(b+a)=0
(a+b)*(a-2b)=0
则a=-b或a=2b
a/b-b/a-(a^2+b^2)/ab
=(a^2-b^2-a^2-b^2)/ab
=-2(b^2)/ab
=-2(b/a)
a=-b或a=2b代入
=2或-1
收起
晕死,大学生来做初中生的题目。。。