Question

Giải phương trình: \(\dfrac{1}{3x^2}+\dfrac{1}{x^2-8x+32}=\dfrac{1}{x^2-2x+8}\)

Answer 1

Nhớ tick cho mình nha

\(\dfrac{1}{3}\)x² + \(\dfrac{1}{x^2}\) - 8x + 32 = \(\dfrac{1}{x^2}\) - 2x + 8  ĐK: x ≠ 0

⇔\(\dfrac{1}{3}\)x² + \(\dfrac{1}{x^2}\) - \(\dfrac{1}{x^2}\) - 8x + 2x + 32 - 8 = 0

⇔\(\dfrac{1}{3}\)x² - 6x +24 = 0

⇔\(\left(x-12\right)\) \(\left(x-6\right)\) = 0

⇔\(\left[{}\begin{matrix}x-12=0\\x-6=0\end{matrix}\right.\)

⇔\(\left[{}\begin{matrix}x=12\left(tm\right)\\x=6\left(tm\right)\end{matrix}\right.\)

⇒ S = \(\left\{12;6\right\}\)