Question

解方程X-1分之X减X-2分之X-1=X-4分之X-3减X-5分之X-4

参考答案：x/(x-1)-(x-1)/(x-2)=(x-3)/(x-4)-(x-4)/(x-5), (x-1+1)/(x-1)-(x-2+1)/(x-2)=(x-4+1)/(x-4)-(x-5+1)/(x-5), 1+1/(x-1)-1-1/(x-2)=1+1/(x-4)-1-1/(x-5), 1/(x-1)-1/(x-2)=1/(x-4)-1/(x-5), 1/(x-1)+1/(x-5)=1/(x-4)+1/(x-2), (x-5+x-1)/[(x-1)(x-5)]=(x-2+x-4)/[(x-4)(x-2)], (2x-6)/(x^2-6x+5)=(2x-6)/(x^2-6x+8), (2x-6)[1/(x^2-6x+5)-1/(x^2-6x+8)]=0, (2x-6)(x^2-6x+8-x^2+6x-5)/[(x^2-6x+5)(x^2-6x+8)]=0, 3(2x-6)/[(x^2-6x+5)(x^2-6x+8)]=0, 2x-6=0, x=3. 不得照抄参考答案！！！！！！！！ 我不是他舅，我告诉你顾客是上帝！还有 看看其他人的答案！他们都太懒了！就会动嘴皮子！我才不会给他们分呢！ 我又有新题目，网站是：http://zhidao.baidu.com/question/142599527.html?fr=middle_ask

Answer 1

x/(x-1)-(x-1)/(x-2)=(x-3)/(x-4)-(x-4)/(x-5)
x/(x-1)+(x-4)/(x-5)=(x-1)/(x-2)+(x-3)/(x-4)
[x/(x-1)-1]+[(x-4)/(x-5)-1]=[(x-1)/(x-2)-1]+[(x-3)/(x-4)-1]
1/(x-1)+1/(x-5)=1/(x-2)+1/(x-4)
通分
(x-1+x-5)/(x-1)(x-5)=(x-2+x-4)/(x-2)(x-4)
(2x-6)/(x²-6x+5)-(2x-6)/(x²-6x+8)=0
(2x-6)[1/(x²-6x+5)-1/(x²-6x+8)]=0
x²-6x+5≠x²-6x+8
所以1/(x²-6x+5)-1/(x²-6x+8)≠0
所以2x-6=0
x=3
分式方程要检验
经检验，x=3是方程的解

我从来不抄别人
你不满意拉倒

Answer 2

x/(x-1)-(x-1)/(x-2)=(x-3)/(x-4)-(x-4)/(x-5), (x-1+1)/(x-1)-(x-2+1)/(x-2)=(x-4+1)/(x-4)-(x-5+1)/(x-5), 1+1/(x-1)-1-1/(x-2)=1+1/(x-4)-1-1/(x-5), 1/(x-1)-1/(x-2)=1/(x-4)-1/(x-5), 1/(x-1)+1/(x-5)=1/(x-4)+1/(x-2), (x-5+x-1)/[(x-1)(x-5)]=(x-2+x-4)/[(x-4)(x-2)], (2x-6)/(x^2-6x+5)=(2x-6)/(x^2-6x+8), (2x-6)[1/(x^2-6x+5)-1/(x^2-6x+8)]=0, (2x-6)(x^2-6x+8-x^2+6x-5)/[(x^2-6x+5)(x^2-6x+8)]=0, 3(2x-6)/[(x^2-6x+5)(x^2-6x+8)]=0, 2x-6=0, x=3.

Answer 3

a²=1-b²>=0
所以0<=b²<=1

2a²+7b²=2(1-b²)+7b²
=5b²+2

0<=b²<=1
0<=5b²<=5
2<=5b²+2<=7
所以最小值=2

Answer 4

用程序写出来就可以了，没难度。

Answer 5

就是三，过程有点多，但并不难

Answer 6

/(x-1)-(x-1)/(x-2)=(x-3)/(x-4)-(x-4)/(x-5), (x-1+1)/(x-1)-(x-2+1)/(x-2)=(x-4+1)/(x-4)-(x-5+1)/(x-5), 1+1/(x-1)-1-1/(x-2)=1+1/(x-4)-1-1/(x-5), 1/(x-1)-1/(x-2)=1/(x-4)-1/(x-5), 1/(x-1)+1/(x-5)=1/(x-4)+1/(x-2), (x-5+x-1)/[(x-1)(x-5)]=(x-2+x-4)/[(x-4)(x-2)], (2x-6)/(x^2-6x+5)=(2x-6)/(x^2-6x+8), (2x-6)[1/(x^2-6x+5)-1/(x^2-6x+8)]=0, (2x-6)(x^2-6x+8-x^2+6x-5)/[(x^2-6x+5)(x^2-6x+8)]=0, 3(2x-6)/[(x^2-6x+5)(x^2-6x+8)]=0, 2x-6=0, /(x-1)-(x-1)/(x-2)=(x-3)/(x-4)-(x-4)/(x-5)
x/(x-1)+(x-4)/(x-5)=(x-1)/(x-2)+(x-3)/(x-4)
[x/(x-1)-1]+[(x-4)/(x-5)-1]=[(x-1)/(x-2)-1]+[(x-3)/(x-4)-1]
1/(x-1)+1/(x-5)=1/(x-2)+1/(x-4)