Question

7/(x+1)-(x+4)/(2-2x)=(3x^2-38)/(x^2-1) помогите срочно 15 баллов

Answer 1

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