7/(x+1)-(x+4)/(2-2x)=(3x^2-38)/(x^2-1

Question

7/(x+1)-(x+4)/(2-2x)=(3x^2-38)/(x^2-1

Answer 1

7/(x+1)-(x+4)/(2-2x)=(3x²-38)/(x²-1)    ОДЗ: x+1≠0  x≠-1   2-2x≠0   2x≠2   x≠1.
7/(x+1)+(x+4)/(2*(x-1))=(3x²-38)/(x²-1)
(7*2*(x-1)+(x+4)*(x+1)/(2*(x²-1))=(3x²-38)/(x²-1)
(14x-14+x²+5x+4)/(x²-1)=2*(3x²-38)/(x²-1)
x²+19x-10=6x²-76
5x²-19x-66=0   D=1681   √D=41
x₁=6   x₂=-2,2
Ответ: x₁=6     x₂=-2,2.

Answer 2

Одз x≠1 x≠-1



D=361+1320=1681=41^2

x₁=(19+41)/10=6
x₂=(19-41)/10=-22/10=-2.2