B=(1/3+3/x^2-3x) : (x^2/27-3x^2+1/2-x) a)tim dk xd b)RG B c)tìm x nguyên để B nguyên

0 bình luận về “B=(1/3+3/x^2-3x) : (x^2/27-3x^2+1/2-x) a)tim dk xd b)RG B c)tìm x nguyên để B nguyên”

1. Đáp án:

   $\begin{array}{l}
   B = \left( {\frac{1}{3} + \frac{3}{{{x^2} – 3x}}} \right):\left( {\frac{{{x^2}}}{{27 – 3{x^2}}} + \frac{1}{{3 + x}}} \right)\\
   a)Đkxd:\left\{ \begin{array}{l}
   {x^2} – 3x \ne 0\\
   27 – 3{x^2} \ne 0\\
   3 + x \ne 0
   \end{array} \right. \Rightarrow \left\{ \begin{array}{l}
   x\left( {x – 3} \right) \ne 0\\
   3\left( {9 – {x^2}} \right) \ne 0\\
   x \ne  – 3
   \end{array} \right. \Rightarrow \left\{ \begin{array}{l}
   x \ne 0;x \ne 3\\
   x \ne 3;x \ne  – 3\\
   x \ne  – 3
   \end{array} \right. \Rightarrow x \ne  – 3;x \ne 0;x \ne 3\\
   b)B = \left( {\frac{1}{3} + \frac{3}{{{x^2} – 3x}}} \right):\left( {\frac{{{x^2}}}{{27 – 3{x^2}}} + \frac{1}{{3 + x}}} \right)\\
    = \left( {\frac{1}{3} + \frac{3}{{x\left( {x – 3} \right)}}} \right):\left( {\frac{{{x^2}}}{{3\left( {3 – x} \right)\left( {3 + x} \right)}} + \frac{1}{{3 + x}}} \right)\\
    = \frac{{{x^2} – 3x + 9}}{{3x\left( {x – 3} \right)}}:\left( {\frac{{{x^2} + 3\left( {3 – x} \right)}}{{3\left( {3 – x} \right)\left( {3 + x} \right)}}} \right)\\
    = \frac{{{x^2} – 3x + 9}}{{3x\left( {x – 3} \right)}}.\frac{{3\left( {3 – x} \right)\left( {3 + x} \right)}}{{{x^2} – 3x + 9}}\\
    = \frac{{ – x – 3}}{x} =  – 1 – \frac{3}{x}\\
   c)B \in Z\\
    \Rightarrow  – 1 – \frac{3}{x} \in Z\\
    \Rightarrow \frac{3}{x} \in Z\\
    \Rightarrow x \in Ư\left( 3 \right) = {\rm{\{ }}1;3\} 
   \end{array}$

   Vậy x=1 hoặc x=3 thì B nguyên

   Bình luận