\[\begin{array}{l}
P = \frac{{4x}}{{3\left( {{x^2} - x + 1} \right)}}\\
a)P = \frac{8}{9}\\
\Leftrightarrow \frac{{4{\rm{x}}}}{{3\left( {{x^2} - x + 1} \right)}} = \frac{8}{9}\\
\Leftrightarrow 36{\rm{x}} = 24\left( {{x^2} - x + 1} \right)\\
\Leftrightarrow 36{\rm{x}} = 24{{\rm{x}}^2} - 24{\rm{x}} + 24\\
\Leftrightarrow 24{{\rm{x}}^2} - 60{\rm{x}} + 24 = 0\\
\Leftrightarrow 24{{\rm{x}}^2} - 48{\rm{x - 12x}} + 24 = 0\\
\Leftrightarrow 24{\rm{x}}\left( {x - 2} \right) - 12\left( {x - 2} \right) = 0\\
\Leftrightarrow 12\left( {x - 2} \right)\left( {2{\rm{x}} - 1} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x - 2 = 0\\
2{\rm{x}} - 1 = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 2\\
x = \frac{1}{2}
\end{array} \right.\\
Vay:x \in \{ 2;\frac{1}{2}\} thi:P = \frac{8}{9}
\end{array}\]

b) Chưa rõ đề

P=4x3(x2−x+1)a)P=89⇔4x3(x2−x+1)=89⇔36x=24(x2−x+1)⇔36x=24x2−24x+24⇔24x2−60x+24=0⇔24x2−48x−12x+24=0⇔24x(x−2)−12(x−2)=0⇔12(x−2)(2x−1)=0⇔[x−2=02x−1=0⇔[x=2x=12Vay:x∈{2;12}thi:P=89P=4x3(x2−x+1)a)P=89⇔4x3(x2−x+1)=89⇔36x=24(x2−x+1)⇔36x=24x2−24x+24⇔24x2−60x+24=0⇔24x2−48x−12x+24=0⇔24x(x−2)−12(x−2)=0⇔12(x−2)(2x−1)=0⇔[x−2=02x−1=0⇔[x=2x=12Vay:x∈{2;12}thi:P=89

b) Chưa rõ đề

P=4x3(x2−x+1)a)P=89⇔4x3(x2−x+1)=89⇔36x=24(x2−x+1)⇔36x=24x2−24x+24⇔24x2−60x+24=0⇔24x2−48x−12x+24=0⇔24x(x−2)−12(x−2)=0⇔12(x−2)(2x−1)=0⇔[x−2=02x−1=0⇔[x=2x=12Vay:x∈{2;12}thi:P=89P=4x3(x2−x+1)a)P=89⇔4x3(x2−x+1)=89⇔36x=24(x2−x+1)⇔36x=24x2−24x+24⇔24x2−60x+24=0⇔24x2−48x−12x+24=0⇔24x(x−2)−12(x−2)=0⇔12(x−2)(2x−1)=0⇔[x−2=02x−1=0⇔[x=2x=12Vay:x∈{2;12}thi:P=89

b) Chưa rõ đề