\(\frac{2}{x^2-4}-\frac{x-1}{x\left(x-2\right)}+\frac{x-4}{x\left(x+2\right)}=0\)Đk \(x\ne\pm2;x\ne0\)
\(\Rightarrow\frac{2}{\left(x-2\right)\left(x+2\right)}-\frac{x-1}{x\left(x-2\right)}+\frac{x-4}{x\left(x+2\right)}=0\)
\(\Rightarrow\frac{2x-\left(x-1\right)\left(x+2\right)+\left(x-4\right)\left(x-2\right)}{x\left(x-2\right)\left(x+2\right)}=0\)
\(\Rightarrow2x-\left(x-1\right)\left(x+2\right)+\left(x-4\right)\left(x-2\right)=0\)
\(\Rightarrow2x-x^2-x+2+x^2-6x+8=0\)
\(\Rightarrow-5x+10=0\)
\(\Rightarrow-5x=-10\)
\(\Rightarrow x=2\)Loại
Ko có gt x thỏa mãn
\(\frac{1}{3-x}-\frac{1}{x+1}=\frac{x}{x-3}-\frac{\left(x-1\right)^2}{x^2-2x-3}\)
\(\Rightarrow\frac{1}{3-x}-\frac{1}{x+1}=\frac{x}{x-3}-\frac{\left(x-1\right)^2}{x^2-3x+x-3}\)
\(\Rightarrow\frac{1}{3-x}-\frac{1}{x+1}=\frac{x}{x-3}-\frac{\left(x-1\right)^2}{\left(x-3\right)\left(x+1\right)}\)Đk \(x\ne3;x\ne-1\)
\(\Leftrightarrow\frac{1}{3-x}-\frac{1}{x+1}-\frac{x}{x-3}-\frac{\left(x-1\right)^2}{\left(x-3\right)\left(x+1\right)}=0\)
\(\Rightarrow-\frac{1}{x-3}-\frac{1}{x+1}-\frac{x}{x-3}+\frac{\left(x-1\right)^2}{\left(x-3\right)\left(x+1\right)}=0\)
\(\Rightarrow\frac{-1\left(x+1\right)-1\left(x-3\right)-x\left(x+1\right)+\left(x-1\right)^2}{\left(x-3\right)\left(x+1\right)}=0\)
\(\Rightarrow-\left(x+1\right)-\left(x-3\right)-x\left(x+1\right)+\left(x-1\right)^2=0\)
\(\Rightarrow x-1-x+3-x^2-x+x^2-2x+1=0\)
\(\Rightarrow-3x+3=0\)
\(\Rightarrow-3x=-3\)
\(\Rightarrow x=1\)
