Đáp án:
\(\begin{array}{l}
4) - 1\\
5)\dfrac{{a + b}}{{a - b}}
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
4)DK:a \ne b;a \ne 0;b \ne 0\\
\dfrac{2}{{ab}}:{\left( {\dfrac{{b - a}}{{ab}}} \right)^2} - \dfrac{{{a^2} + {b^2}}}{{{{\left( {a - b} \right)}^2}}}\\
= \dfrac{2}{{ab}}.\dfrac{{{{\left( {ab} \right)}^2}}}{{{{\left( {a - b} \right)}^2}}} - \dfrac{{{a^2} + {b^2}}}{{{{\left( {a - b} \right)}^2}}}\\
= \dfrac{{2ab - {a^2} - {b^2}}}{{{{\left( {a - b} \right)}^2}}}\\
= \dfrac{{ - {{\left( {a - b} \right)}^2}}}{{{{\left( {a - b} \right)}^2}}} = - 1\\
5)DK:a \ne 0;b \ne 0;a \ne b\\
\left[ {\dfrac{a}{{b\left( {a - b} \right)}} - \dfrac{b}{{a\left( {a - b} \right)}}} \right].\dfrac{{ab\left( {a + b} \right)}}{{\left( {a + b} \right)\left( {a - b} \right)}}\\
= \dfrac{{{a^2} - {b^2}}}{{ab\left( {a - b} \right)}}.\dfrac{{ab\left( {a + b} \right)}}{{\left( {a + b} \right)\left( {a - b} \right)}}\\
= \dfrac{{\left( {a + b} \right)\left( {a - b} \right)}}{{ab\left( {a - b} \right)}}.\dfrac{{ab\left( {a + b} \right)}}{{\left( {a + b} \right)\left( {a - b} \right)}}\\
= \dfrac{{a + b}}{{a - b}}
\end{array}\)
