Question

Giải phương trình: \(\dfrac{1}{x-2}+3=\dfrac{3-x}{x-2}\)

Answer 1

\(\dfrac{1}{x-2}+3=\dfrac{3-x}{x-2}\left(1\right)\)
ĐK: \(x-2\ne0\Leftrightarrow x\ne2\)
\(\left(1\right)\Leftrightarrow\dfrac{1}{x-2}+\dfrac{3\left(x-2\right)}{x-2}=\dfrac{3-x}{x-2}\)
\(\Rightarrow1+3\left(x-2\right)=3-x\)
\(\Leftrightarrow1+3x-6=3-x\)
\(\Leftrightarrow3x+x=-1+6+3\)
\(\Leftrightarrow4x=8\)
\(\Leftrightarrow x=2\left(L\right)\)
Vậy \(S=\varnothing\)

Answer 2

\(\dfrac{1}{x-2}+3=\dfrac{3-x}{x-2}\)

\(\Leftrightarrow\dfrac{1}{x-2}+\dfrac{3\left(x-2\right)}{x-2}=\dfrac{3-x}{x-2}\)

\(\Leftrightarrow1+3x-6=3-x\)

\(\Leftrightarrow4x=8\)

\(\Leftrightarrow x=2\)