$\frac{x+1}{x-1}$ -$\frac{4}{x+1}$ +$\frac{x^2-3}{1-x^2}$ =0 đk:x khác ±1
⇔$\frac{x+1}{x-1}$ -$\frac{4}{x+1}$ +$\frac{x^2-3}{(1-x)(1+x)}$ =0
⇔$\frac{x+1}{x-1}$ -$\frac{4}{x+1}$ -$\frac{x^2-3}{(x+1)(x-1)}$ =0
⇔$\frac{(x+1)(x+1)-4(x-1)-x^2+3}{(x+1)(x-1)}$=0
⇔$\frac{x^2+2x+1-4x+4-x^2+3}{(x+1)(x-1)}$=0
⇔$\frac{-2x+8}{(x+1)(x-1)}$=0 ⇔-2x+8=0
⇔x=4
Vậy S=4
hoặc là Vậy x=4
b)$\frac{2}{x^2+2x+1}$ -$\frac{5}{x^2-2x+1}$ =$\frac{3}{1-x^2}$
⇔$\frac{2}{x^2+2x+1}$ -$\frac{5}{x^2-2x+1}$ -$\frac{3}{1-x^2}$ =0
⇔$\frac{2}{(x+1)^2}$ -$\frac{5}{(x-1)^2}$ -$\frac{3}{(1-x)(1+x)}$ =0
⇔$\frac{2}{(1+x)^2}$ -$\frac{5}{(1-x)^2}$ -$\frac{3}{(1-x)(1+x)}$ =0
⇔$\frac{2(1-x)^2-5(1+x)^2-3(1-x)(1+x)}{(1+x)^2.(1-x)^2}$=0
⇔2(1-x)^2-5(1+x)^2-3(1-x)(1+x)=0
⇔2-4x+2x^2-5-10x-5x^2-3+3x^2=0
⇔-14x-6=0
⇔x=-$\frac{3}{7}$
(kết quả này đg tính lại bạn đừng tin tg quá nhé :/ )
c)$\frac{x+1}{x^2+x+1}$ -$\frac{x-1}{x^2-x+1}$ =$\frac{x}{x(x^4+x^2+1)}$
⇔$\frac{(x+1)(x^2-2x+1)-(x-1)(x^2+2x+1)}{(x^2+x+1)(x^2-2x+1)}$=$\frac{x}{x(x^4+x^2+1)}$
⇔$\frac{x^3+1-x^3+1}{x(x^4+x^2+1)}$ =$\frac{x}{x(x^4+x^2+1)}$
⇔x^3+1-x^3+1=x
⇔2=x
⇔x=2
Vậy x=2
