Question

giải pt

\(\dfrac{2}{x-1}\)>1

Answer 1

\(\dfrac{2}{x-1}>1\)

\(\Leftrightarrow\dfrac{2}{x-1}-1>0\)

\(\Leftrightarrow\dfrac{2-x+1}{x-1}>0\)

\(\Leftrightarrow\dfrac{3-x}{x-1}>0\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}3-x>0\\x-1>0\end{matrix}\right.\\\left\{{}\begin{matrix}3-x< 0\\x-1< 0\end{matrix}\right.\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x< 3\\x>1\end{matrix}\right.\\\left\{{}\begin{matrix}x>3\\x< 1\end{matrix}\right.\end{matrix}\right.\)\(\Leftrightarrow1< x< 3\)

:")))))))

Answer 2

À bổ xung thêm cái đkxđ: \(x\ne1\)

Answer 3

\(\dfrac{2}{x-1}>1\) ( x # 1)

⇔ \(\dfrac{2}{x-1}-1>0\)

⇔ \(\dfrac{3-x}{x-1}>0\)

lập bảng xét dấu , ta có :

Vậy , nghiệm của BPT : 1 < x < 3