Đáp án:

Giải thích các bước giải:

\(\begin{array}{l}
a)\quad \left(x + \dfrac1x\right)^2 +2\left(x + \dfrac1x\right) - 8 =0\qquad (ĐK:x \ne 0)\\
Đặt\,\,t = x + \dfrac1x\qquad (t \geq 2)\\
\text{Phương trình trở thành:}\\
\quad t^2 + 2t - 8 =0\\
\Leftrightarrow (t+4)(t-2) =0\\
\Leftrightarrow \left[\begin{array}{l}t = -4\quad (loại)\\t = 2\quad (nhận)\end{array}\right.\\
Với\,t =2\,\,ta\,\,được:\\
\quad x+ \dfrac1x = 2\\
\Leftrightarrow x^2 - 2x + 1 =0\\
\Leftrightarrow (x-1)^2 =0\\
\Leftrightarrow x = 1 \quad (nhận)\\
Vậy\,\,x = 1\\
b)\quad \left(x - \dfrac1x\right)^2 -3\left(x - \dfrac1x\right)+\dfrac89 =0\qquad (ĐK: x \ne 0)\\
\Leftrightarrow x^2 - 2 + \dfrac{1}{x^2} - 3x + \dfrac3x + \dfrac89 =0\\
\Leftrightarrow (x^2 - 3x) + \dfrac{3x+1}{x^2} - \dfrac{10}{9} =0\\
\Leftrightarrow (x^2 -3x) + \dfrac{27x + 9 - 10x^2}{9x^2} =0\\
\Leftrightarrow x(x-3) - \dfrac{(10x+3)(x-3)}{9x^2} =0\\
\Leftrightarrow \dfrac{(x-3)(3x+1)(3x^2-x-3)}{9x^2}=0\\
\Leftrightarrow \left[\begin{array}{l}x = 3\\x = -\dfrac13\\3x^2 - x - 3 =0\end{array}\right.\\
\Leftrightarrow \left[\begin{array}{l}x = 3\\x = -\dfrac13\\\left(x-\dfrac16\right)^2 = \dfrac{37}{36}\end{array}\right.\\
\Leftrightarrow \left[\begin{array}{l}x = 3\\x = -\dfrac13\\x = \dfrac16 + \dfrac{\sqrt{37}}{36}\\x = \dfrac16 - \dfrac{\sqrt{37}}{36}\end{array}\right.\\
c)\quad (x+2)(x^2 -10)(x-2)=40\\
\Leftrightarrow (x^2 - 4)(x^2 - 10) = 40\\
Đặt\,\,t = x^2 - 4\\
\text{Phương trình trở thành:}\\
\quad t(t-6) = 40\\
\Leftrightarrow (t+4)(t-10) =0\\
\Leftrightarrow \left[\begin{array}{l}t =-4\\t = 10\end{array}\right.\\
+)\quad Với\,t=-4\,\,ta\,\,được:\\
\quad x^2 - 4 = -4\\
\Leftrightarrow x^2 = 0\\
\Leftrightarrow x =0\\
+)\quad Với\,t =10\,\,ta\,\,được:\\
\quad x^2 - 4 = 10\\
\Leftrightarrow x^2 = 14\\
\Leftrightarrow x = \pm \sqrt{14}\\
Vậy\,\,x = 0\,\,hoặc\,\,x = \pm \sqrt{14}\\
d)\quad (x^2 - 4x)^2 + (x-2)^2 = 10\\
\Leftrightarrow (x^2 - 4x)^2 + x^2 - 4x - 6 =0\\
Đặt\,\,t = x^2 - 4x\\
\text{Phương trình trở thành:}\\
\quad t^2 + t - 6 = 0\\
\Leftrightarrow (t-2)(t+3) =6\\
\Leftrightarrow \left[\begin{array}{l}t = 2\\t =-3\end{array}\right.\\
+)\quad \,t = 2\,\,ta\,\,được:\\
\quad x^2 - 4x = 2\\
\Leftrightarrow (x-2)^2 = 6\\
\Leftrightarrow \left[\begin{array}{l}x - 2 = \sqrt6\\x - 2 = -\sqrt6\end{array}\right.\\
\Leftrightarrow \left[\begin{array}{l}x=2+ \sqrt6\\x =2 -\sqrt6\end{array}\right.\\
+)\quad \,t = -3\,\,ta\,\,được:\\
\quad x^2 - 4x = -3\\
\Leftrightarrow x^2 - 4x + 3 =0\\
\Leftrightarrow (x-1)(x-3) =0\\
\Leftrightarrow \left[\begin{array}{l}x = 1\\x = 3\end{array}\right.\\
Vậy\,\,x = 1,\, x = 3\, hoặc\,x = 2 \pm \sqrt6
\end{array}\)