Question

\(\frac{x-2}{1-x}=\frac{1}{x-1}-1-2x\)

Answer 1

ĐKXĐ: x≠1

Ta có: \(\frac{x-2}{1-x}=\frac{1}{x-1}-1-2x\)

\(\Leftrightarrow\frac{x-2}{1-x}-\frac{-1}{1-x}+1+2x=0\)

\(\Leftrightarrow\frac{x-2}{1-x}-\frac{-1}{1-x}+\frac{1-x}{1-x}+\frac{2x\left(1-x\right)}{1-x}=0\)

\(\Leftrightarrow x-2+1+1-x+2x-2x^2=0\)

\(\Leftrightarrow-2x^2+2x=0\)

\(\Leftrightarrow2x-2x^2=0\)

\(\Leftrightarrow2x\left(1-x\right)=0\)

Vì 2≠0

nên \(\left[{}\begin{matrix}x=0\\1-x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\left(ktm\right)\end{matrix}\right.\Leftrightarrow x=0\)

Vậy: x=0

Answer 2

\(\frac{x-2}{1-x}\)=\(\frac{1}{x-1}\)-1-2x (ĐKXĐ:x≠1)

⇔\(\frac{2-x}{x-1}\)=\(\frac{1}{x-1}\)-\(\frac{x-1}{x-1}\)-\(\frac{2x\left(x-1\right)}{x-1}\)

⇒2-x=1-x+1-2x²+2x

⇔2-x-1+x-1+2x²-2x=0

⇔2x²-2x=0

⇔2x(x-1)=0

⇔2x=0 hoặc x-1=0

⇔x=0 hoặc x=1(KTMĐKXĐ)

Vậy tập nghiệm của phương trình đã cho là:S={0}