Question

Giải phương trình: \(\frac{x-2}{x-3}\) + \(\frac{x+2}{x-5}\) = \(\frac{2}{\left(x-3\right)\left(5-x\right)}\)

Answer 1

\(\frac{x-2}{x-3}+\frac{x+2}{x-5}=\frac{2}{\left(x-3\right)\left(5-x\right)}\)

\(\Leftrightarrow\) \(\frac{\left(x-2\right)\left(x-5\right)}{\left(x-3\right)\left(x-5\right)}+\frac{\left(x+2\right)\left(x-3\right)}{\left(x-5\right)\left(x-3\right)}+\frac{2}{\left(x-3\right)\left(x-5\right)}=0\)

\(\Leftrightarrow\) (x - 2)(x - 5) + (x + 2)(x - 3) + 2 = 0

\(\Leftrightarrow\) x² - 5x - 2x + 10 + x² - 3x + 2x - 6 + 2 = 0

\(\Leftrightarrow\) 2x² - 8x + 6 = 0

\(\Leftrightarrow\) 2(x² - 4x + 3) = 0

\(\Leftrightarrow\) x² - 4x + 3 = 0

\(\Leftrightarrow\) x = 1

Vậy S = {1}

Chúc bn học tốt!!