Question

Giải pt:

\(\frac{x+5}{x-1}=\frac{x+1}{x-3}-\frac{8}{x^2-4x+3}\)

Answer 1

\(\frac{x+5}{x-1}=\frac{x+1}{x-3}-\frac{8}{x^2-4x+3}\) (ĐKXĐ: x \(\ne\) 1; x \(\ne\) 3)

\(\Leftrightarrow\) \(\frac{\left(x+5\right)\left(x-3\right)}{\left(x-1\right)\left(x-3\right)}-\frac{\left(x+1\right)\left(x-1\right)}{\left(x-1\right)\left(x-3\right)}+\frac{8}{\left(x-1\right)\left(x-3\right)}=0\)

\(\Rightarrow\) (x + 5)(x - 3) - (x + 1)(x - 1) + 8 = 0

\(\Leftrightarrow\) x² - 3x + 5x - 15 - x² + 1 + 8 = 0

\(\Leftrightarrow\) 2x - 6 = 0

\(\Leftrightarrow\) 2x = 6

\(\Leftrightarrow\) x = 3

Vậy S = {3}

Chúc bn học tốt!!