Question

$\frac{4}{ x -1} - \frac{4}{x + 1} = 1$
Помогите с алгеброй пожалуйста, срочно

Answer 1

$\frac{4}{x-1}-\frac{4}{x+1}=1\; \; ,\qquad ODZ:\; x\ne -1\; ,\; x\ne 1\\\\\frac{4(x+1)-4(x-1)-(x-1)(x+1)}{(x-1)(x+1)}=0\\\\\frac{4x+4-4x+4-x^2+1}{(x-1)(x+1)}=0\\\\\frac{-x^2+9}{(x-1)(x+1)}=0\; \; \Rightarrow \; \; \left \{ {{-x^2+9=0} \atop {x\ne 1\; ,\; x\ne -1}} \right. \\\\-x^2+9=0\; \; ,\; \; x^2-9=0\; \; ,\\\\(x-3)9x+3)=0\\\\\underline {x_1=3\; ,\; \; x_2=-3}$

Answer 2

4/(x-1)-4/(x+1)=1 , x≠1,x≠-1

(4(x+1)-4(x-1))/(x-1)(x+1) = 1

(4x+4-4x+4)/(x-1)(x+1)=1

8/(x²-1)=1

x²-1=8

x²=8+1

x²=9

a)/x/=3, x1=3, x2=-3

b)x²-9=0, (x+3)(x-3)=0

  x-3=0, x1=3

 x+3=0, x2=-3