( $\sqrt[]{x+1}$ – $\sqrt[]{x-2}$ )(1+ $\sqrt[]{x^2-x-2}$) = 3

0 bình luận về “( $\sqrt[]{x+1}$ – $\sqrt[]{x-2}$ )(1+ $\sqrt[]{x^2-x-2}$) = 3”

1. Đáp án:

   x=3

   Giải thích các bước giải:

   \(\begin{array}{l}
   DK:\,x \ge 2\\
   Dat\,\sqrt {x + 1}  = a;\sqrt {x – 2}  = b\,\,\left( {a,b \ge 0} \right)\\
    \Rightarrow {a^2} – {b^2} = 3\\
   Ta\,co\,hpt\\
   \left\{ \begin{array}{l}
   \left( {a – b} \right)\left( {1 + ab} \right) = 3\\
   {a^2} – {b^2} = 3
   \end{array} \right.\\
    \Rightarrow \left( {a – b} \right)\left( {a + b} \right) – \left( {a – b} \right)\left( {1 + ab} \right) = 0\\
    \Leftrightarrow \left( {a – b} \right)\left( {a + b – ab – 1} \right) = 0\\
    \Leftrightarrow \left( {a – b} \right)\left( {a – 1 – b\left( {a – 1} \right)} \right) = 0\\
    \Leftrightarrow \left( {a – b} \right)\left( {a – 1} \right)\left( {b – 1} \right) = 0\\
    \Leftrightarrow \left[ \begin{array}{l}
   a = b\\
   a = 1\\
   b = 1
   \end{array} \right.\\
   TH1:\,a = b \Leftrightarrow \sqrt {x + 1}  = \sqrt {x – 2} \left( {VN} \right)\\
   TH2:\,a = 1 \Rightarrow {b^2} =  – 2\left( {VN} \right)\\
   TH3:\,b = 1 \Rightarrow a = 2\\
   Suy\,ra\,\left\{ \begin{array}{l}
   \sqrt {x + 1}  = 2\\
   \sqrt {x – 2}  = 1
   \end{array} \right. \Leftrightarrow x = 3\left( {tm} \right)
   \end{array}\)

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