\(\dfrac{x-1}{x-3}>1\)
⇔ \(\dfrac{x-1}{x-3}>\dfrac{x-3}{x-3}\)
⇔ `x-1>x-3`
⇔ x-x>1-3
⇔ 0x>-2 (luôn đúng)
Vậy ....
ĐKXĐ \(x\ne3\)
\(\dfrac{x-1}{x-3}>1\Leftrightarrow\dfrac{x-1}{x-3}-1>0\Leftrightarrow\dfrac{x-1-x+3}{x-3}>0\Leftrightarrow\dfrac{2}{x-3}>0\Leftrightarrow x-3>0\Leftrightarrow x>3\)
Vậy, x > 3 thì bpt thỏa mãn
\(\dfrac{x-1}{x-3}>1 \)
\(\dfrac{x-1}{x-3}-1>0\)
\(\dfrac{x-1}{x-3}-\dfrac{x-3}{x-3}>0\)
\(\dfrac{2}{x-3}>0\) (tử mẫu cùng dấu) mà 2>0
=>x-3>0
x>3