gpt: a) `(x+1/x)^2+2(x+1/x)-8=0` b) `(x-1/x)^2-3(x-1/x)+8/9=0` c) `(x+2)(x^2-10)(x-2)=40` d) `(x^2-4x)^2+(x-2)^2=10`

0 bình luận về “gpt: a) `(x+1/x)^2+2(x+1/x)-8=0` b) `(x-1/x)^2-3(x-1/x)+8/9=0` c) `(x+2)(x^2-10)(x-2)=40` d) `(x^2-4x)^2+(x-2)^2=10`”

1. 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}\)

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