Đáp án:
\({\dfrac{{\left( { - 2{x^2} + 12x} \right)\left( {{x^3} + 3{x^2} + 3x + 2} \right)}}{{\left( {9 - {x^2}} \right)\left( {8 - x} \right)}}}\)
Giải thích các bước giải:
\(\begin{array}{*{20}{l}}
{DK:x \ne \pm 3;x \ne 8}\\
{A = \left( {\dfrac{{3 + x}}{{3 - x}} - \dfrac{{3 - x}}{{x + 3}} + \dfrac{{4{x^2}}}{{{x^2} - 9}}} \right).\left( {\left( {2 + x} \right).\dfrac{{{x^2} + x + 1}}{{8 - x}}} \right)}\\
{ = \left[ {\dfrac{{{x^2} + 6x + 9 - 9 + 6x - {x^2} - 4{x^2}}}{{\left( {3 - x} \right)\left( {x + 3} \right)}}} \right].\left[ {\dfrac{{2{x^2} + 2x + 2 + {x^3} + {x^2} + x}}{{8 - x}}} \right]}\\
{ = \dfrac{{ - 2{x^2} + 12x}}{{\left( {3 - x} \right)\left( {x + 3} \right)}}.\dfrac{{{x^3} + 3{x^2} + 3x + 2}}{{8 - x}}}\\
{ = \dfrac{{\left( { - 2{x^2} + 12x} \right)\left( {{x^3} + 3{x^2} + 3x + 2} \right)}}{{\left( {9 - {x^2}} \right)\left( {8 - x} \right)}}}
\end{array}\)
\(\begin{array}{l}
b.2x = 4\\
\to x = 2\\
\to A = \dfrac{{\left( { - 2.4 + 12.2} \right)\left( {8 + 3.4 + 3.2 + 2} \right)}}{{\left( {9 - 4} \right)\left( {8 - 2} \right)}}\\
= \dfrac{{16.28}}{{5.6}} = \dfrac{{224}}{{15}}
\end{array}\)
