a,
P = $\frac{x}{x-1}$ + $\frac{3}{x+1}$ - $\frac{6x-4}{x^{2}-1}$
P = $\frac{x(x+1)}{(x-1)(x+1)}$ + $\frac{3(x-1)}{(x+1)(x-1)}$ - $\frac{6x-4}{(x-1)(x+1)}$
P = $\frac{x^2+x}{(x-1)(x+1)}$ + $\frac{3x-3}{(x-1)(x+1)}$ - $\frac{6x-4}{(x-1)(x+1)}$
P = $\frac{x^2+x+3x-3-6x+4}{(x-1)(x+1)}$
P = $\frac{x^2-2x+1}{(x-1)(x+1)}$
P = $\frac{(x-1)^2}{(x-1)(x+1)}$
P = $\frac{x-1}{x+1}$
Vậy P = $\frac{x-1}{x+1}$
b,
P = $\frac{1}{2}$
⇔ $\frac{x-1}{x+1}$ = $\frac{1}{2}$
⇔ $\frac{2(x-1)}{2(x+1)}$ = $\frac{x+1}{2(x+1)}$
⇔ $\frac{2x-2}{2(x+1)}$ = $\frac{x+1}{2(x+1)}$
⇒ 2x - 2 = x + 1
⇔ 2x - x = 1 + 2
x = 3 (tm ĐKXĐ)
Vậy khi x = 3 thì P = $\frac{1}{2}$
c,
P > 1
⇔ $\frac{x-1}{x+1}$ > 1
⇔ $\frac{x-1}{x+1}$ > $\frac{x+1}{x+1}$
⇒ x - 1 > x + 1
⇔ x - x > 1+1
⇔ 0 > 2 (vô lý)
Vậy không tìm được x để P >1.
