\(\dfrac{x-4}{x+4}-\dfrac{x}{x-4}=\dfrac{3x-14}{x^2-16}\)
\(\Leftrightarrow\dfrac{x-4}{x+4}-\dfrac{x}{x-4}=\dfrac{3x-14}{\left(x-4\right)\left(x+4\right)}\)
ĐKXĐ:
\(x+4\ne0\Leftrightarrow x\ne-4\)
\(x-4\ne0\Leftrightarrow x\ne4\)
\(\dfrac{x-4}{x+4}-\dfrac{x}{x-4}=\dfrac{3x-14}{\left(x-4\right)\left(x+4\right)}\)
\(\Leftrightarrow\left(x-4\right)\left(x-4\right)-x\left(x+4\right)=3x-14\)
\(\Leftrightarrow x^2-4x-4x+16-x^2-4x-3x+14=0\)
\(\Leftrightarrow-15x+30=0\)
\(\Leftrightarrow-15x=-30\)
\(\Leftrightarrow x=2\)(nhận)
Vậy \(S=\left\{2\right\}\)
ĐKXĐ: x ≠ 4; x ≠ -4
Phương trình đã cho tương đương:
(x - 4)² - x(x + 4) = 3x - 14
⇔ x² - 8x + 16 - x² - 4x =3x - 14
⇔ -8x - 4x - 3x = -14 - 16
⇔ -15x = -30
⇔ x = 2 (nhận)
Vậy S = {2}
