Đáp án:
Giải thích các bước giải:
\(\begin{array}{*{20}{l}}
{A = \left( {\frac{1}{3} + \frac{3}{{{x^2} - 3x}}} \right):\left( {\frac{{{x^2}}}{{27 - 3{x^2}}} + \frac{1}{{x + 3}}} \right)}\\
{a.DK:x \ne 0;x \ne {\rm{ \;}} \pm 3}\\
{A = \frac{{3{x^2} - 9x + 9}}{{3x\left( {x - 3} \right)}}:\frac{{ - {x^2} + 3\left( {x - 3} \right)}}{{3\left( {x + 3} \right)\left( {x - 3} \right)}}}\\
{ = \frac{{{x^2} - 3x + 3}}{{x\left( {x - 3} \right)}}.\frac{{3\left( {x + 3} \right)\left( {x - 3} \right)}}{{ - {x^2} + 3x - 9}}}\\
{ = \frac{{(3{x^2} - 9x + 9)(x + 3)}}{{x\left( { - {x^2} + 3x - 9} \right)}}}\\
{b.A < {\rm{ \;}} - 1}\\
{ \to \frac{{(3{x^2} - 9x + 9)(x + 3)}}{{x\left( { - {x^2} + 3x - 9} \right)}} < {\rm{ \;}} - 1}\\
{ \to \frac{{3{x^3} - 18x + 27 - {x^3} - 3{x^2} + 9x}}{{x\left( { - {x^2} - 3x + 9} \right)}} < 0}\\
{ \to \frac{{2{x^3} - 3{x^2} - 9x + 27}}{{x\left( { - {x^2} - 3x + 9} \right)}} < 0\left( * \right)}
\end{array}\)
