Ответы
Решение во вложении:
![](https://st.uroker.com/files/27d/27d98c74080e1396ba6ba6761eb40f56.jpg)
X/(x²-x) + 2/(1-x²) =5/(x²+x) ODZ: ,x²-x=0 i x²+x =0
X(x-1)=0 x(x+1)=0
X=0 x-1=0 x+1=0
X=0 x=1 x=-1
x/x(x-1) +2/-(x²-1) =5/ x(x+1)
1/(x-1) -2/(x-1)(x+1)=5/x(x+1)
X(x+1) – 2x -5(x-1) / x(x+1)(x-1 =0
(X²+x-2x-5x+5) / (x(x²-1) =0 // * x(x²-1)
X²-6x+5=0
D=36-20=16 , √D=4
X1=(6-4)/2=2/2=1 неt корнем уравнения, входит в ОДЗ
X2=(6+4)/2=10/2=5
Проверка:
5/(5²-5) +2/(1-5²) = 5/(5²+5
5/(25-5) + 2/(1-25)= 5/(25+5)
5/20 +2/-24 = 5/30
1/4 - 1/12 = 1/6
(3-1) /12 =1/6
2/12=1/6
1/6=1/6
![](https://st.uroker.com/files/27d/27d98c74080e1396ba6ba6761eb40f56.jpg)