Question

giải phương trình:

\(\left(\dfrac{x-1}{x}\right)^2\)-\(3\left(x-\dfrac{1}{x}\right)\)+\(\dfrac{8}{9}\)=0

Answer 1

Giờ lười rồi, xin slot mai giải cho nhé :>

Answer 2

\(\left(x-\dfrac{1}{x}\right)^2-3\left(x-\dfrac{1}{x}\right)+\dfrac{8}{9}=0\)

=>x-1/x=8/3 hoặc x-1/x=1/3

=>\(\left[{}\begin{matrix}\dfrac{x^2-1}{x}=\dfrac{8}{3}\\\dfrac{x^2-1}{x}=\dfrac{1}{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x^2-3=8x\\3x^2-3=x\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x^2-8x-3=0\\3x^2-x-3=0\end{matrix}\right.\)

=>\(x\in\left\{3;-\dfrac{1}{3};\dfrac{1+\sqrt{37}}{6};\dfrac{1-\sqrt{37}}{6}\right\}\)