\(\frac{x}{x-1}-\frac{1}{1-x}-1=2\) ĐKXĐ : \(x\ne1\)
\(\Leftrightarrow\frac{x}{x-1}+\frac{1}{x-1}-\frac{x-1}{x-1}=\frac{2\left(x-1\right)}{x-1}\)
\(\Rightarrow x+1-x+1=2x-2\)
\(\Leftrightarrow x+1-x+1-2x+2=0\)
\(\Leftrightarrow-2x+4=0\)
\(\Leftrightarrow-2x=-4\)
\(\Leftrightarrow x=2\left(tm\text{đ}k\right)\)
Vậy tập nghiệm của phương trình là S = { 2 } .
\(\frac{x}{x-1}-\frac{1}{1-x}-1=2\) ( ĐKXĐ : \(x\ne1\))
\(\Leftrightarrow\frac{x}{x-1}+\frac{1}{x-1}-\frac{x-1}{x-1}=\frac{2\left(x-1\right)}{x-1}\)
\(\Rightarrow x+1-x+1=2x-2\)
\(\Rightarrow x+1-x+1-2x+2=0\)
\(\Rightarrow-2x+4=0\)
\(\Rightarrow-2x=-4\)
\(\Rightarrow x=2\left(tm\right)\)
Vậy phương trình có nghiệm duy nhất là x = 2
\(\frac{x}{x-1}-\frac{1}{1-x-1}=2\)
\(\frac{x}{x-1}+\frac{1}{x}=2\)
\(\frac{x^2}{x\left(x-1\right)}+\frac{x-1}{x\left(x-1\right)}=2\)
\(\Leftrightarrow\frac{x^2+x-1}{x\left(x-1\right)}=2\)
\(\Leftrightarrow x^2+x-1=2\left(x-1\right)x\)
\(\Leftrightarrow x^2+x-1=2x^2-2x\)
\(\Leftrightarrow x^2-2x^2+x+2x-1=0\)
\(\Leftrightarrow-x^2+3x-1=0\)
\(\Leftrightarrow x^2-3x+1=0\)
\(\Leftrightarrow x^2-2x+1-x=0\)
\(\Leftrightarrow\left(x-1\right)^2-\left(\sqrt{x}\right)^2=0\)
\(\Leftrightarrow\left(x-1-\sqrt{x}\right)\left(x-1+\sqrt{x}\right)=0\)
tự làm
