Rút gọn biểu thức sau: A=(1/3+3/x^-3x):(x^/27-3x^+1/x+3)

0 bình luận về “Rút gọn biểu thức sau: A=(1/3+3/x^-3x):(x^/27-3x^+1/x+3)”

1. Đáp án:

   \(\dfrac{{2{x^3} + 54}}{{ – 3{x^3} + 6{x^2} – 18x}}\)

   Giải thích các bước giải:

   \(\begin{array}{l}
   DK:x \ne \left\{ { – 3;0;3} \right\}\\
   A = \left( {\dfrac{1}{3} + \dfrac{3}{{{x^2} – 3x}}} \right):\left( {\dfrac{{{x^2}}}{{27 – 3{x^2}}} + \dfrac{1}{{x + 3}}} \right)\\
    = \left[ {\dfrac{{{x^2} – 3x + 3.3}}{{3x\left( {x – 3} \right)}}} \right]:\left[ {\dfrac{{{x^2}}}{{2\left( {9 – {x^2}} \right)}} + \dfrac{1}{{x + 3}}} \right]\\
    = \left[ {\dfrac{{{x^2} – 3x + 9}}{{3x\left( {x – 3} \right)}}} \right]:\left[ {\dfrac{{{x^2} + 2\left( {3 – x} \right)}}{{2\left( {3 – x} \right)\left( {3 + x} \right)}}} \right]\\
    = \left[ {\dfrac{{{x^2} – 3x + 9}}{{3x\left( {x – 3} \right)}}} \right].\dfrac{{2\left( {3 – x} \right)\left( {3 + x} \right)}}{{{x^2} – 2x + 6}}\\
    = \dfrac{{{x^2} – 3x + 9}}{{ – 3x}}.\dfrac{{2\left( {x + 3} \right)}}{{{x^2} – 2x + 6}}\\
    = \dfrac{{{x^2} – 3x + 9}}{{ – 3x}}.\dfrac{{2x + 6}}{{{x^2} – 2x + 6}}\\
    = \dfrac{{2{x^3} + 6{x^2} – 6{x^2} – 18x + 18x + 54}}{{ – 3{x^3} + 6{x^2} – 18x}}\\
    = \dfrac{{2{x^3} + 54}}{{ – 3{x^3} + 6{x^2} – 18x}}
   \end{array}\)

   Trả lời