Đáp án:
a) \(\dfrac{{2x}}{{x - 3}}\)
b) P=-1
Giải thích các bước giải:
\(\begin{array}{l}
a)P = \dfrac{{{{\left( {x - 2} \right)}^2} + x\left( {x + 2} \right) - 2x - 4}}{{\left( {x - 2} \right)\left( {x + 2} \right)}}.\dfrac{{x - 3 + 5}}{{x - 3}}\\
= \dfrac{{{x^2} - 4x + 4 + {x^2} + 2x - 2x - 4}}{{\left( {x - 2} \right)\left( {x + 2} \right)}}.\dfrac{{x + 2}}{{x - 3}}\\
= \dfrac{{2{x^2} - 4x}}{{\left( {x - 2} \right)\left( {x + 2} \right)}}.\dfrac{{x + 2}}{{x - 3}}\\
= \dfrac{{2x\left( {x - 2} \right)}}{{\left( {x - 2} \right)\left( {x + 2} \right)}}.\dfrac{{x + 2}}{{x - 3}}\\
= \dfrac{{2x}}{{x - 3}}\\
b){x^2} - 3x + 2 = 0\\
\to \left( {x - 2} \right)\left( {x - 1} \right) = 0\\
\to \left[ \begin{array}{l}
x = 2\left( l \right)\\
x = 1
\end{array} \right.\\
Thay:x = 1\\
\to P = \dfrac{{2.1}}{{1 - 3}} = - 1
\end{array}\)
