Question

解方程，需步骤⑴（2x-5）/（x^2-3x+2）+4/（x^2-4）=1/（x-2)

⑵1/(X+7)=（（x+1）/（2x-1）(x+7)）+(1/（2x^2-3x+1）)

Answer 1

⑴（2x-5）/（x^2-3x+2）+4/（x^2-4）=1/（x-2)
(2x-5）/（x-2)(x-1)+4/（x+2)(x-2)=1/（x-2)
(2x-5）/(x-1)+4/（x+2)=1
(2x-5)(x+2)+4(x-1)=(x-1)(x+2)
2x^2+4x-5x-10+4x-4=x^2+2x-x-2
x^2+2x-12=0
用公式法解得
x=(-2+√2^2+4*1*12)/2=-1+√13
或x=(-2-√2^2+4*1*12)/2=-1-√13
⑵1/(X+7)=（（x+1）/（2x-1）(x+7)）+(1/（2x^2-3x+1）)
1/(X+7)=（（x+1）/（2x-1）(x+7)）+(1/（2x-1)(x-1))
1=（x+1）/（2x-1）+(x+7)/（2x-1)(x-1)
（2x-1)(x-1)=（x+1）（x-1）+(x+7)
2x^2-3x+1=x^2-1+x+7
x^2-4x-5=0
(x-5)(x+1)=0
x=5或x=-1
经验证x=5或x=-1是方程的解。

Answer 2

1/（2x^2-3x+1）)看不懂这是什么