初二数学分式问题。80分悬赏。。详细一点。

2025-03-29 12:24:52
推荐回答(4个)
回答1:

1.原式变为 (x-m^2)/(x-3)=2
所以x=6-m^2
又从方程中可知,x不等于3
所以当6-m^2=3时,该方程无解
即 m为正负根号3

2.
1/a-1/b=4,
(b-a)/(ab)=4,
b-a=4ab,
a-b=-4ab,
(a-2ab-b)/(2a-2b+7ab)
=[(a-b)-2ab]/[2(a-b)+7ab]
=(-4ab-2ab)/(-8ab+7ab)
=(-6ab)/(-ab)
=6.

3.
x/(x-3)-2=m/(x-3)
[x-2(x-3)]/(x-3)=m/(x-3)

(6-x)/(x-3)=m/(x-3)
因为x-3不等于0
所以6-x=m
x=6-m>0
m<6

4.
因为解为0,所以jiangx=0代入,得-1/2=(2a-3)/(a+5)

-a-5=4a-6

a=1/5

5.
1/(4-x^2)+2=k/(x-2)
-1/(x+2)(x-2)+2=k/(x-2)
两边乘(x+2)(x-2)

-1+2(x+2)(x-2)=k(x+2)
分式方程的增根就是分母为0
即(x+2)(x-2)=0
x=2,x=-2

x=2,代入-1+2(x+2)(x-2)=k(x+2)
-1=4k,k=-1/4

x=-2,代入-1+2(x+2)(x-2)=k(x+2)
-1=0,不成立

所以k=-1/4

回答2:

1.解分式方程,得x=6-m²,因为方程无解,所以出现了增根,所以x=3,所以6-m²=3,解得m=±根号3。
2.把1/a-1/b=4通分,得(b-a)/ab=4,也就是a-b=-4ab。代数式a-2ab-b/2a-2b+7ab=(a-b)-2ab/2(a-b)+7ab=-4ab-2ab/2*(-4ab)+7ab=-6ab/-ab=6。
3.解这个方程,得x=6-m,因为有一个正数解,所以不能出现增根,也就是x>0,且x≠3,所以6-m>0,且6-m≠3,解得m<6且m≠3。
4.把x=0代入方程,得-1/2=2a-3/a+5,交叉相乘,得-(a+5)=2(2a-3),所以-a-5=4a-6,解得a=1/5。
5.把这个方程化为整式方程,得-9-2x²=kx+2k,因为方程有增根,所以x=2或-2。
当x=2时,把它代入方程,得-9-2*2²=2k+2k,-9-8=4k,k=-17/4。
当x=-2时,把它代入方程,得-9-2*(-2)²=-2k+2k,方程无解。
所以k=-17/4。

回答3:

(1)通分化简得(6-x)/(x-3)=m² /(x-3).若无解,则是因为解是x=3使分母无意义。所以m=正负根号3
(2)上下同除以ab得{-(1/a-1/b)-2}/{-2(1/a-1/b)+7}=-6/-1=6
(3)同一题。6-x=m且x不等于3.x》0.所以m<6且不等于3
(4)若解是x=0.带入则2a-3/a+5=-2然后求解就好
(5)方程两边同乘x²-4得2x²-kx-(2k+9)=0.有增根说明x=±2.分别带入得k=-1/4

回答4:

分别为正负根号三、6、正负根号六之间、五分之一、负四分之一

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