Question

giải phương trình

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

Answer 1

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

<=> \(\frac{1}{x-2}+\frac{3\left(x-2\right)}{x-2}+\frac{x-3}{x-2}=0\)

<=> \(\frac{1+3x-6+x-3}{x-2}=0\)

<=> \(4x-8=0\)

<=> 4x = 8

<=> x = 2

Vậy S = {2}

Answer 2

\(\Leftrightarrow\) \(\frac{1}{x-2}\) + 3 = \(\frac{3-x}{x-2}\)
\(\Leftrightarrow\) \(\frac{1}{x-2}\) + \(\frac{3\left(x-2\right)}{x-2}\) = \(\frac{3-x}{x-2}\)
\(\Leftrightarrow\) 1 + 3x - 6 = 3 - x
\(\Leftrightarrow\) 3x + x = 3 - 1 + 6
\(\Leftrightarrow\) 4x = 8
\(\Leftrightarrow\) x = 2
Vậy S = {2}