Đáp án:
$x+\dfrac{1}{x}=\dfrac{5}{2}$
$\text{ĐKXĐ : x $\neq$ 0 }$
$⇔\dfrac{2x^2}{2x} + \dfrac{2}{2x} = \dfrac{5x}{2x}$
$⇔2x^2+2=5x$
$⇔2x^2-5x+2=0$
$⇔ 2x^2-4x-x+2=0$
$⇔2x(x-2)-(x-2)=0$
$⇔(x-2)(2x-1)=0$
⇔\(\left[ \begin{array}{l}x-2=0\\2x-1=0\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=2(TM)\\x=\dfrac{1}{2}(TM)\end{array} \right.\)
$a) x^2 + \dfrac{1}{x^2}$
$\text{-Thay $x=2$ vào biểu thức, ta có : }$
$2^2 + \dfrac{1}{2^2}$
$=4 +\dfrac{1}{4}$
$=\dfrac{16}{4}+\dfrac{1}{4}$
$=\dfrac{17}{4}$
$\text{-Thay $x=\dfrac{1}{2}$ vào biểu thức, ta có : }$
$(\dfrac{1}{2})^2 + 1 : (\dfrac{1}{2})^2$
$=\dfrac{1}{4} + 1 :\dfrac{1}{4}$
$=\dfrac{1}{4} + 4$
$=\dfrac{1}{4} + \dfrac{16}{4}$
$=\dfrac{17}{4}$
$b) x^4 + \dfrac{1}{x^4}$
$\text{-Thay x=2 vào biểu thức, ta có : }$
$2^4 +\dfrac{1}{2^4}$
$=16+\dfrac{1}{16}$
$=\dfrac{256}{16} + \dfrac{1}{6}$
$=\dfrac{257}{16}$
$\text{-Thay $x=\dfrac{1}{2}$ vào biểu thức, ta có : }$
$(\dfrac{1}{2})^4 + 1 : (\dfrac{1}{4})^2$
$=\dfrac{1}{16} + 1 : \dfrac{1}{16}$
$=\dfrac{1}{16} + 16$
$=\dfrac{1}{16} + \dfrac{256}{16}$
$=\dfrac{257}{16}$
