x −1x−3>1
x−1x−3- 1 >0
x−1x−3- \(\dfrac{\text{x − 3}}{\text{x − 3}}\) > 0
\(\dfrac{-4}{\text{x − 3}}\) > 0
=> -4 và x-3 khác dấu
mà -4 <0
=> x-3 > 0
=> x . 3
x−1x−3>1
Ta có: \(\dfrac{x-1}{x-3}>1\)
\(\Leftrightarrow\dfrac{x-1}{x-3}-1>0\)
\(\Leftrightarrow\dfrac{x-1-x+3}{x-3}>0\)
\(\Leftrightarrow\dfrac{2}{x-3}>0\)
mà 2>0
nên x-3>0
hay x>3