\(\dfrac{1}{x-2}+3=\dfrac{x-3}{2-x}\)
ĐKXĐ: x khác 2
\(\Leftrightarrow\)\(\dfrac{1}{x-2}+3-\dfrac{x-3}{2-x}=0\)
\(\Leftrightarrow\) \(\dfrac{1+3\cdot\left(x-2\right)+x-3}{x-2}=0\)
\(\Leftrightarrow\) 1 + 3x - 6 + x - 3 = 0
\(\Leftrightarrow\) 4x - 8 = 0
\(\Leftrightarrow\) 4x = 8
\(\Leftrightarrow\) x = 2 (KTMĐKXĐ)
Vậy pt vô nghiệm