作业帮求数学题目两道!
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求数学题目两道!
求数学题目两道!
求数学题目两道!
A(x-1)+B(x+1)=x-3
(A+B)x+B-A=x-3
A+B=1
B-A=-3
B=-1
A=2
1 首先等式两边同乘以（x+1）（x-1） 原题就变为x-3=A（x-1）+B（x+1） =（A+B）x+B-A
也就是A+B=1 A-B=3 解出A=2 B=-1
分母化成一样的（忘记叫啥了）
a(x-1)+b(x+1)=x-3
a+b=1
b-a=-3
a=2
b=-1
A/(x+1)+B/(x-1)=[A(x-1)+B(x+1)]/(x+1)(x-1)=[(A+B)x+(B-a)]/(x+1)(x-1) A+B=1 (1) B-A=-3(2)
(1)+(2)2B=-2 B=-1 A=1-B=2
y/(x-y)-x^2y^2/(x^3-x^2y)(x^2-y^2)/(xy^2+y^3=y/(x-y)-x^2y^2/(x-y)(x+y)x^2y^2(x-y(x+y)=y/(x-y)-1
x=1/2 y=-1 原式=y/(x-y)-1=-2/3-1=-5/3
(x-3)/[(x-1)(x+1)]=A/(x+1)+B/(x-1) 通分
(x-3)/[(x-1)(x+1)]=[A(x-1)+B(x+1)]/[(x+1)(x-1)]
(x-3)/[(x-1)(x+1)]=[Ax-A+Bx+B]/[(x+1)(x-1)]
(x-3)/[(x-1)(x+1)]=[(A+B)x-(A-B)]/[(x+1)(x-1)]
x-3...
全部展开
(x-3)/[(x-1)(x+1)]=A/(x+1)+B/(x-1) 通分
(x-3)/[(x-1)(x+1)]=[A(x-1)+B(x+1)]/[(x+1)(x-1)]
(x-3)/[(x-1)(x+1)]=[Ax-A+Bx+B]/[(x+1)(x-1)]
(x-3)/[(x-1)(x+1)]=[(A+B)x-(A-B)]/[(x+1)(x-1)]
x-3=(A+B)x-(A-B)
A+B=1
A-B=3
两式相加得
2A=4
A=2
两式相减得
2B=-2
B=-1
y/(x-y)-x²y²/(x³-x²y)*(x²-y²)/(xy²+y³)
=y/(x-y)-x²y²/[x²(x-y)]*[(x-y)(x+y)]/(xy²+y³)
=y/(x-y)-y²*(x+y)]/[y²(x+y)]
=y/(x-y)-(x+y)/(x+y)
=y/(x-y)-1
=-1/[1/2-(-1)]-1
=-1/(3/2)-1
=-2/3-1
=-5/3
收起