Question

解方程1(/x2+x-2)+1/(x2+7x+10)=2

要具体过程

Answer 1

1/(x^2+x-2)+1/(x^2+7x+10)=2
1/[(x+2)(x-1)]+1/[(x+5)(x+2)]=2
1/(x-1)+1/(x+5)=2(x+2)
2(x+2)/[(x-1)(x+5)=2(x+2)
当x=-2时 分母为0，无意义，即x+2≠0
则(x-1)(x+5)=1
x^2+4x-6=0
(x+2)=10
x=-2±√10

Answer 2

左边=1/[(x+2)(x-1)]+1/[(x+2)(x+5)] ：有公因式1/(x+2)
=[1/(x+2)] [1/(x-1) +1/(x+5)] ：右边中括号内同分
=[1/(x+2)] * (x-1+x+5)/[(x-1)(x+5)]
=[1/(x+2)] * 2(x+2)/[(x-1)(x+5)] ：可消去因式(x+2)
=2/[(x-1)(x+5)]
所以(x-1)(x+5)=1
x^2+4x=6
(x+2)^2=10
得x=-2±（根号10）

Answer 3

1/(x-1)(x+2)+1/(x+2)(x+5)=2
(x+5)+(x-1)=2(x-1)(x+2)(x+5)
2x+4=2(x-1)(x+2)(x+5)
x+2=(x-1)(x+2)(x+5)
[(x-1)(x+5)-1](x+2)=0
(x^2+4x-5-1)(x+2)=0
(x^2+4x-6)(x+2)=0
x=-2
x^2+4x-6=0
（x+2）^-10=0
x+2=±√10
x=-2±√10

Answer 4

1/(x-1)(x+2)+1/(x+2)(x+5)=2
(x+5)+(x-1)=2(x-1)(x+2)(x+5)
2x+4=2(x-1)(x+2)(x+5)
x+2=(x-1)(x+2)(x+5)
[(x-1)(x+5)-1](x+2)=0
(x^2+4x-5-1)(x+2)=0
(x^2+4x-6)(x+2)=0
所以x=-2或x^2+4x-6=0(公式法，过程略)

Answer 5

这是什么，发错了？