Giải bpt: căn(6x^2 – 18x + 12) < 3x + 10 - x^2

0 bình luận về “Giải bpt: căn(6x^2 – 18x + 12) < 3x + 10 - x^2”

1. Đáp án:

   \(x \in \left( { – 1;1} \right] \cup \left[ {2;4} \right)\)

   Giải thích các bước giải:

   \(\begin{array}{l}
   \sqrt {6{x^2} – 18x + 12}  < 3x + 10 – {x^2}\\
    \to \sqrt {6\left( {{x^2} – 3x + 2} \right)}  + {x^2} – 3x – 10 < 0\left( 1 \right)\\
   Đặt:\sqrt {6\left( {{x^2} – 3x + 2} \right)}  = t\left( {t \ge 0} \right)\\
    \to {t^2} = 6\left( {{x^2} – 3x + 2} \right)\\
    \to {x^2} – 3x + 2 = \dfrac{{{t^2}}}{6}\\
   \left( 1 \right) \to t + \dfrac{{{t^2}}}{6} – 12 < 0\\
    \to {t^2} + 6t – 72 < 0\\
    \to \left( {t – 6} \right)\left( {t + 12} \right) < 0\\
    \to t \in \left( { – 12;6} \right)\\
    \to t \in \left[ {0;6} \right)\\
    \to \left\{ \begin{array}{l}
   \sqrt {6\left( {{x^2} – 3x + 2} \right)}  \ge 0\\
   \sqrt {6\left( {{x^2} – 3x + 2} \right)}  < 6
   \end{array} \right.\\
    \to \left\{ \begin{array}{l}
   {x^2} – 3x + 2 \ge 0\\
   6\left( {{x^2} – 3x + 2} \right) < 36
   \end{array} \right.\\
    \to \left\{ \begin{array}{l}
   \left( {x – 2} \right)\left( {x – 1} \right) \ge 0\\
   {x^2} – 3x + 2 – 6 < 0
   \end{array} \right.\\
    \to \left\{ \begin{array}{l}
   x \in \left( { – \infty ;1} \right] \cup \left[ {2; + \infty } \right)\\
   \left( {x – 4} \right)\left( {x + 1} \right) < 0
   \end{array} \right.\\
    \to \left\{ \begin{array}{l}
   x \in \left( { – \infty ;1} \right] \cup \left[ {2; + \infty } \right)\\
   x \in \left( { – 1;4} \right)
   \end{array} \right.\\
   KL:x \in \left( { – 1;1} \right] \cup \left[ {2;4} \right)
   \end{array}\)

   Trả lời