Question

\(2\left(x^2+\dfrac{1}{x^2}\right)-x-\dfrac{1}{x}-6=0\)

Answer 1

ĐKXĐ: \(x\ne0\)

Ta có: \(2\left(x^2+\dfrac{1}{x^2}\right)-x-\dfrac{1}{x}-6=0\)

\(\Leftrightarrow2\left(x^2+\dfrac{1}{x^2}+2-2\right)-x-\dfrac{1}{x}-6=0\)

\(\Leftrightarrow2\left(x+\dfrac{1}{x}\right)^2-\left(x+\dfrac{1}{x}\right)-10=0\)

\(\Leftrightarrow2\left(x+\dfrac{1}{x}\right)^2+4\left(x+\dfrac{1}{x}\right)-5\left(x+\dfrac{1}{x}\right)-10=0\)

\(\Leftrightarrow2\left(x+\dfrac{1}{x}\right)\left(x+\dfrac{1}{x}+2\right)-5\left(x+\dfrac{1}{x}+2\right)=0\)

\(\Leftrightarrow\left(x+\dfrac{1}{x}+2\right)\left(2x+\dfrac{2}{x}-5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{x}+2=0\\2x+\dfrac{2}{x}-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\dfrac{x^2}{x}+\dfrac{1}{x}+\dfrac{2x}{x}=0\\\dfrac{2x^2+2-5x}{x}=0\end{matrix}\right.\)

Suy ra: \(\left[{}\begin{matrix}x^2+2x+1=0\\2x^2-5x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left(x+1\right)^2=0\\\left(x-2\right)\left(2x-1\right)=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\x-2=0\\2x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\left(nhận\right)\\x=2\left(nhận\right)\\x=\dfrac{1}{2}\left(nhận\right)\end{matrix}\right.\)

Vậy: \(S=\left\{-1;2;\dfrac{1}{2}\right\}\)

Answer 2

2(\(x^2+\dfrac{1}{x^2}\)) - (x + \(\dfrac{1}{x}\)) - 6 = 0

đặt x + \(\dfrac{1}{x}\) = a

<=>2(a² - 2) - a - 6 = 0

<=> 2a² - a - 10 = 0

<=> (2a - 5)(a + 2) = 0

TH1: a = \(\dfrac{5}{2}\)

=> x + \(\dfrac{1}{x}\) = \(\dfrac{5}{2}\) 

<=> x² + 1 = \(\dfrac{5x}{2}\) 

<=> 2x² - 5x + 1 = 0

<=> x₁ = \(\dfrac{5+\sqrt{17}}{4}\) ;x₂ = \(\dfrac{5-\sqrt{17}}{4}\)

TH2: a = -2

=> x + \(\dfrac{1}{x}\) = -2

<=> x² + 2x + 1 = 0

<=> (x + 1)² = 0

<=> x + 1 = 0

<=> x = -1

S = {\(\dfrac{5+\sqrt{17}}{4}\);\(\dfrac{5-\sqrt{17}}{4}\);-1}