Question

4 пример решите пожалуйста,даю 40 баллов!!!!!

Answer 1

Ответ:

$\frac{{5302}}{{729}}$

Объяснение:

$\displaystyle\frac{{{a^{12}} + 1}}{{{a^6}}} = {a^6} + \displaystyle\frac{1}{{{a^6}}} = {({a^2})^3} + \displaystyle\frac{1}{{{{({a^2})}^3}}} =\\\\=\left( {{a^2} + \displaystyle\frac{1}{{{a^2}}}} \right)\left( {{a^4} - {a^2} \cdot \displaystyle\frac{1}{{{a^2}}} + \displaystyle\frac{1}{{{a^4}}}} \right) = \left( {{a^2} + \displaystyle\frac{1}{{{a^2}}}} \right)\left( {{a^4} + \displaystyle\frac{1}{{{a^4}}} - 1} \right).$

Отдельно посчитаем значения выражений

${a^2} + \displaystyle\frac{1}{{{a^2}}}$ и ${a^4} + \displaystyle\frac{1}{{{a^4}}}.$

$a - \displaystyle\frac{1}{a} = \displaystyle\frac{2}{3};\\\\{\left( {a - \displaystyle\frac{1}{a}} \right)^2} = {\left( {\displaystyle\frac{2}{3}} \right)^2};\\\\{a^2} - 2 \cdot a \cdot \displaystyle\frac{1}{a} + \displaystyle\frac{1}{{{a^2}}} = \displaystyle\frac{4}{9};$

${a^2} + \displaystyle\frac{1}{{{a^2}}} = \displaystyle\frac{{22}}{9}.{\left( {{a^2} + \displaystyle\frac{1}{{{a^2}}}} \right)^2} = {\left( {\displaystyle\frac{{22}}{9}} \right)^2};\\\\{a^4} + 2{a^2} \cdot \displaystyle\frac{1}{{{a^2}}} + \displaystyle\frac{1}{{{a^4}}} = \displaystyle\frac{{484}}{{81}};\\\\{a^4} + \displaystyle\frac{1}{{{a^4}}} = \displaystyle\frac{{322}}{{81}}.$

Значит

$\displaystyle\frac{{{a^{12}} + 1}}{{{a^6}}} = \left( {{a^2} + \displaystyle\frac{1}{{{a^2}}}} \right)\left( {{a^4} + \displaystyle\frac{1}{{{a^4}}} - 1} \right) = \displaystyle\frac{{22}}{9} \cdot \left( {\displaystyle\frac{{322}}{{81}} - 1} \right) = \displaystyle\frac{{22}}{9} \cdot \displaystyle\frac{{241}}{{81}} = \displaystyle\frac{{5302}}{{729}}.$

Answer 2

Відповідь: фото

Пояснення: