已知对于x不等于3且x不等于-2的所有x的值,等式x的平方-x-6分之3x+1=x-3分之A+( x+2)分之B总能成立.求A,B值已知y=x-i分之x-1的绝对值-(x-2)的绝对值分之x-2+ (3x-x的平方-2)分之x的平方-3x+2的绝
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![已知对于x不等于3且x不等于-2的所有x的值,等式x的平方-x-6分之3x+1=x-3分之A+( x+2)分之B总能成立.求A,B值已知y=x-i分之x-1的绝对值-(x-2)的绝对值分之x-2+ (3x-x的平方-2)分之x的平方-3x+2的绝](/uploads/image/z/1043957-29-7.jpg?t=%E5%B7%B2%E7%9F%A5%E5%AF%B9%E4%BA%8Ex%E4%B8%8D%E7%AD%89%E4%BA%8E3%E4%B8%94x%E4%B8%8D%E7%AD%89%E4%BA%8E-2%E7%9A%84%E6%89%80%E6%9C%89x%E7%9A%84%E5%80%BC%2C%E7%AD%89%E5%BC%8Fx%E7%9A%84%E5%B9%B3%E6%96%B9-x-6%E5%88%86%E4%B9%8B3x%2B1%3Dx-3%E5%88%86%E4%B9%8BA%2B%EF%BC%88+x%2B2%EF%BC%89%E5%88%86%E4%B9%8BB%E6%80%BB%E8%83%BD%E6%88%90%E7%AB%8B.%E6%B1%82A%2CB%E5%80%BC%E5%B7%B2%E7%9F%A5y%3Dx-i%E5%88%86%E4%B9%8Bx-1%E7%9A%84%E7%BB%9D%E5%AF%B9%E5%80%BC-%EF%BC%88x-2%29%E7%9A%84%E7%BB%9D%E5%AF%B9%E5%80%BC%E5%88%86%E4%B9%8Bx-2%2B+++%EF%BC%883x-x%E7%9A%84%E5%B9%B3%E6%96%B9-2%EF%BC%89%E5%88%86%E4%B9%8Bx%E7%9A%84%E5%B9%B3%E6%96%B9-3x%2B2%E7%9A%84%E7%BB%9D)
已知对于x不等于3且x不等于-2的所有x的值,等式x的平方-x-6分之3x+1=x-3分之A+( x+2)分之B总能成立.求A,B值已知y=x-i分之x-1的绝对值-(x-2)的绝对值分之x-2+ (3x-x的平方-2)分之x的平方-3x+2的绝
已知对于x不等于3且x不等于-2的所有x的值,等式x的平方-x-6分之3x+1=x-3分之A+( x+2)分之B总能成立.求A,B值
已知y=x-i分之x-1的绝对值-(x-2)的绝对值分之x-2+ (3x-x的平方-2)分之x的平方-3x+2的绝对值
已知对于x不等于3且x不等于-2的所有x的值,等式x的平方-x-6分之3x+1=x-3分之A+( x+2)分之B总能成立.求A,B值已知y=x-i分之x-1的绝对值-(x-2)的绝对值分之x-2+ (3x-x的平方-2)分之x的平方-3x+2的绝
1、根据题意
(3x+1)/(x²-x-6)=A/(x-3)+B/(x+2)
(3x+1)/(x²-x-6)=[A(x+2)+B(x-3)]/[(x-3)(x+2)]
(3x+1)/(x²-x-6)=(Ax+2A+Bx-3B)/(x²-x-6)
(3x+1)/(x²-x-6)=[(A+B)x+(2A-3B)]/(x²-x-6)
A+B=3(1)
2A-3B=1(2)
(1)×3+(2)
5A=10
A=2
B=3-A=1
A=2
B=1
2、y=|x-1|/(x-1)-|x-2|/(x-2)+|x²-3x+2|/(3x-x²-2)
=|x-1|/(x-1)-|x-2|/(x-2)-|(x-2)(x-1)|/(x²-3x+2)
=|x-1|/(x-1)-|x-2|/(x-2)-|(x-1)(x-2)|/[(x-1)(x-2)]
零点法
讨论
(1)x
分之A+( x+2)分之B总能成立 ????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????...
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分之A+( x+2)分之B总能成立 ?????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????什么意思
收起
1、根据题意
(3x+1)/(x²-x-6)=A/(x-3)+B/(x+2)
(3x+1)/(x²-x-6)=[A(x+2)+B(x-3)]/[(x-3)(x+2)]
(3x+1)/(x²-x-6)=(Ax+2A+Bx-3B)/(x²-x-6)
(3x+1)/(x²-x-6)=[(A+B)x+(2A-3B)]/(x&...
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1、根据题意
(3x+1)/(x²-x-6)=A/(x-3)+B/(x+2)
(3x+1)/(x²-x-6)=[A(x+2)+B(x-3)]/[(x-3)(x+2)]
(3x+1)/(x²-x-6)=(Ax+2A+Bx-3B)/(x²-x-6)
(3x+1)/(x²-x-6)=[(A+B)x+(2A-3B)]/(x²-x-6)
A+B=3(1)
2A-3B=1(2)
(1)×3+(2)
5A=10
A=2
B=3-A=1
A=2
B=1
2、y=|x-1|/(x-1)-|x-2|/(x-2)+|x²-3x+2|/(3x-x²-2)
=|x-1|/(x-1)-|x-2|/(x-2)-|(x-2)(x-1)|/(x²-3x+2)
=|x-1|/(x-1)-|x-2|/(x-2)-|(x-1)(x-2)|/[(x-1)(x-2)]
零点法
讨论
(1)x<1
x-1<0,x-2<0,(x-1)(x-2)>0
y=(-1)-(-1)-1=-1
(2)1
y=1-(-1)-(-1)=3
(3)x>2
x-1>0,x-2>0,(x-1)(x-2)>0
y=1-1-1=-1
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所以A=2,B=1