Đáp án:
b) \(x = - \dfrac{{19}}{8}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)DK:x \ne \left\{ { - 3;0;3} \right\}\\
P = \left[ {\dfrac{{{{\left( {3 + x} \right)}^2} + 9{x^2} - {{\left( {3 - x} \right)}^2}}}{{\left( {3 - x} \right)\left( {x + 3} \right)}}} \right]:\dfrac{{3x\left( {3x + 4} \right)}}{{{x^2}\left( {3 - x} \right)}}\\
= \dfrac{{9 + 6x + {x^2} + 9{x^2} - 9 + 6x - {x^2}}}{{\left( {3 - x} \right)\left( {x + 3} \right)}}.\dfrac{{{x^2}\left( {3 - x} \right)}}{{3x\left( {3x + 4} \right)}}\\
= \dfrac{{x\left( {9{x^2} + 12x} \right)}}{{3\left( {x + 3} \right)\left( {3x + 4} \right)}}\\
= \dfrac{{3{x^2}\left( {3x + 4} \right)}}{{3\left( {x + 3} \right)\left( {3x + 4} \right)}}\\
= \dfrac{{{x^2}}}{{x + 3}}\\
b)\dfrac{P}{x} - 3P + 19 = 5x\\
\to \dfrac{{{x^2}}}{{x + 3}}:x - 3.\dfrac{{{x^2}}}{{x + 3}} + 19 = 5x\\
\to \dfrac{x}{{x + 3}} - \dfrac{{3{x^2}}}{{x + 3}} + 19 - 5x = 0\\
\to \dfrac{{ - 3{x^2} + x + 19\left( {x + 3} \right) - 5x\left( {x + 3} \right)}}{{x + 3}} = 0\\
\to - 3{x^2} + x + 19\left( {x + 3} \right) - 5x\left( {x + 3} \right) = 0\\
\to - 3{x^2} + x + 19x + 57 - 5{x^2} - 15x = 0\\
\to - 8{x^2} + 5x + 57 = 0\\
\to - 8{x^2} + 24x - 19x + 57 = 0\\
\to - 8x\left( {x - 3} \right) - 19\left( {x - 3} \right) = 0\\
\to \left( {x - 3} \right)\left( { - 8x - 19} \right) = 0\\
\to \left[ \begin{array}{l}
x = 3\left( l \right)\\
x = - \dfrac{{19}}{8}
\end{array} \right.
\end{array}\)
