Đáp án:
$\begin{array}{l}
B1)\\
a)P:\dfrac{{9{x^2} - 4}}{{5x + 3}} = \dfrac{{25{x^2} + 30x + 9}}{{3x - 2}}\\
\Leftrightarrow P = \dfrac{{25{x^2} + 30x + 9}}{{3x - 2}}.\dfrac{{9{x^2} - 4}}{{5x + 3}}\\
= \dfrac{{{{\left( {5x} \right)}^2} + 2.5x.3 + 9}}{{\left( {3x - 2} \right)}}.\dfrac{{\left( {3x - 2} \right)\left( {3x + 2} \right)}}{{5x + 3}}\\
= \dfrac{{{{\left( {5x + 3} \right)}^2}}}{1}.\dfrac{{3x + 2}}{{5x + 3}}\\
= \left( {5x + 3} \right)\left( {3x + 2} \right)\\
= 15{x^2} + 19x + 6\\
b)\dfrac{{x + 1}}{{x + 2}}:\dfrac{{x + 3}}{{x + 2}}:\dfrac{{x + 4}}{{x + 3}}\\
= \dfrac{{x + 1}}{{x + 2}}.\dfrac{{x + 2}}{{x + 3}}.\dfrac{{x + 3}}{{x + 4}}\\
= \dfrac{{x + 1}}{{x + 4}}\\
B2)\\
A = \dfrac{2}{{xy}}:{\left( {\dfrac{1}{x} - \dfrac{1}{y}} \right)^2} - \dfrac{{{x^2} + {y^2}}}{{{x^2} - 2xy + {y^2}}}\\
= \dfrac{2}{{xy}}:\dfrac{{{{\left( {y - x} \right)}^2}}}{{{{\left( {xy} \right)}^2}}} - \dfrac{{{x^2} + {y^2}}}{{{{\left( {x - y} \right)}^2}}}\\
= \dfrac{2}{{xy}}.\dfrac{{{{\left( {xy} \right)}^2}}}{{{{\left( {x - y} \right)}^2}}} - \dfrac{{{x^2} + {y^2}}}{{{{\left( {x - y} \right)}^2}}}\\
= \dfrac{{2xy}}{{{{\left( {x - y} \right)}^2}}} - \dfrac{{{x^2} + {y^2}}}{{{{\left( {x - y} \right)}^2}}}\\
= \dfrac{{{{\left( {x - y} \right)}^2}}}{{{{\left( {x - y} \right)}^2}}}\\
= 1
\end{array}$
Vậy giá trị của A không phụ thuộc của biến
