Задача 15. Упростите выражение    \(\dfrac{{{{\left( {a-b} \right)}^2}}}{a}\left( {\dfrac{a}{{{{\left( {a-b} \right)}^2}}} + \dfrac{a}{{{b^2}-{a^2}}}} \right) + \dfrac{{3a + b}}{{a + b}}\)

Ответ

ОТВЕТ: 3.

Решение

\(\dfrac{{{{\left( {a-b} \right)}^2}}}{a}\left( {\dfrac{a}{{{{\left( {a-b} \right)}^2}}} + \dfrac{a}{{{b^2}-{a^2}}}} \right) + \dfrac{{3a + b}}{{a + b}} = \)

\( = \dfrac{{{{\left( {a-b} \right)}^2}}}{a} \cdot \left( {\dfrac{a}{{{{\left( {a-b} \right)}^2}}}-\dfrac{a}{{\left( {a-b} \right)\left( {a + b} \right)}}} \right) + \dfrac{{3a + b}}{{a + b}} = \)

\( = \dfrac{{{{\left( {a-b} \right)}^2} \cdot \left( {{a^2} + ab-{a^2} + ab} \right)}}{{a{{\left( {a-b} \right)}^2}\left( {a + b} \right)}} + \dfrac{{3a + b}}{{a + b}} = \dfrac{{2ab}}{{a\left( {a + b} \right)}} + \dfrac{{3a + b}}{{a + b}} = \)

\( = \dfrac{{2b}}{{a + b}} + \dfrac{{3a + b}}{{a + b}} = \dfrac{{3a + 3b}}{{a + b}} = \dfrac{{3\left( {a + b} \right)}}{{a + b}} = 3.\)

Ответ:  3.