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

$\frac{7}{x+1} + \frac{x+4}{2(x-1)} = \frac{3x^2-38}{x^2-1} \\ \\ 7(2x-2)+(x+4)(x+1)=2(3x^2-38) \\ 14x-14+x^2+5x+4=6x^2-76 \\ 5x^2-19x-66=0$

D=361+1320=1681=41^2

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