cho bt p=(x-3/x+2-x-2/x+3+2-x/x2+5x+6):(1-x/x-1) a) rút gọn p b)tìm x nguyên để p nguyên c)tìm x để p>1

0 bình luận về “cho bt p=(x-3/x+2-x-2/x+3+2-x/x2+5x+6):(1-x/x-1) a) rút gọn p b)tìm x nguyên để p nguyên c)tìm x để p>1”

1. Đáp án:

   c. \(\left[ \begin{array}{l}
   x >  – 2\\
   x <  – 3
   \end{array} \right.\)

   Giải thích các bước giải:

   \(DK:x \ne \left\{ { – 2; – 3} \right\}\)

   \(\begin{array}{l}
   a.P = \left( {\dfrac{{x – 3}}{{x + 2}} – \dfrac{{x – 2}}{{x + 3}} + \dfrac{{2 – x}}{{{x^2} + 5x + 6}}} \right):\left( {\dfrac{{1 – x}}{{x – 1}}} \right)\\
    = \left[ {\dfrac{{{x^2} – 9 – {x^2} + 4 + 2 – x}}{{\left( {x + 2} \right)\left( {x + 3} \right)}}} \right]:\left( {\dfrac{{1 – x}}{{ – \left( {1 – x} \right)}}} \right)\\
    = \dfrac{{ – 1 – x}}{{\left( {x + 2} \right)\left( {x + 3} \right)}}.\left( { – 1} \right)\\
    = \dfrac{{x + 1}}{{\left( {x + 2} \right)\left( {x + 3} \right)}}\\
   b.P = \dfrac{{x + 1}}{{\left( {x + 2} \right)\left( {x + 3} \right)}} = \dfrac{{x + 2 – 1}}{{\left( {x + 2} \right)\left( {x + 3} \right)}}\\
    = \dfrac{1}{{x + 3}} – \dfrac{1}{{\left( {x + 2} \right)\left( {x + 3} \right)}}\\
   P \in Z\\
    \Leftrightarrow \left\{ \begin{array}{l}
   \dfrac{1}{{x + 3}} \in Z\\
   \dfrac{1}{{\left( {x + 2} \right)\left( {x + 3} \right)}} \in Z
   \end{array} \right.\\
    \Leftrightarrow \left\{ \begin{array}{l}
   x + 3 \in U\left( 1 \right)\\
   {x^2} + 5x + 6 \in U\left( 1 \right)
   \end{array} \right.\\
    \to \left\{ \begin{array}{l}
   \left[ \begin{array}{l}
   x + 3 = 1\\
   x + 3 =  – 1
   \end{array} \right.\\
   \left[ \begin{array}{l}
   {x^2} + 5x + 6 = 1\\
   {x^2} + 5x + 6 =  – 1\left( l \right)
   \end{array} \right.
   \end{array} \right.\\
    \to \left\{ \begin{array}{l}
   \left[ \begin{array}{l}
   x =  – 2\\
   x =  – 4
   \end{array} \right.\\
   \left[ \begin{array}{l}
   x = \dfrac{{ – 5 + \sqrt 5 }}{2}\\
   x = \dfrac{{ – 5 – \sqrt 5 }}{2}
   \end{array} \right.
   \end{array} \right.
   \end{array}\)

   ⇒ Không tồn tại x TMĐK

   \(\begin{array}{l}
   c.P > 1\\
    \to \dfrac{{x + 1}}{{\left( {x + 2} \right)\left( {x + 3} \right)}} > 1\\
    \to \dfrac{{x + 1 – {x^2} – 5x – 6}}{{\left( {x + 2} \right)\left( {x + 3} \right)}} > 0\\
    \to \dfrac{{ – {x^2} – 4x – 5}}{{\left( {x + 2} \right)\left( {x + 3} \right)}} > 0\\
    \to \left( {x + 2} \right)\left( {x + 3} \right) > 0\left( {do: – {x^2} – 4x – 5 < 0\forall x} \right)\\
    \to \left[ \begin{array}{l}
   \left\{ \begin{array}{l}
   x + 2 > 0\\
   x + 3 > 0
   \end{array} \right.\\
   \left\{ \begin{array}{l}
   x + 2 < 0\\
   x + 3 < 0
   \end{array} \right.
   \end{array} \right.\\
    \to \left[ \begin{array}{l}
   x >  – 2\\
   x <  – 3
   \end{array} \right.
   \end{array}\)

   Bình luận