Предмет: Алгебра
Автор: kotovandrej311
Вопрос задан 6 лет назад

CРОЧНО! ПРОШУ
Про число a известно, что a + 1/a = 25/12. Найдите:

a) a^2 + 1/a^2

б) a^3 + 1/a^3

в) a^4 + 1/a^4

Ответы

Ответ дал: Universalka

$\displaystyle\bf\\a+\frac{1}{a} =\frac{25}{12} \\\\\\1)\\\\\Big(a+\frac{1}{a} \Big)^{2} =\Big(\frac{25}{12} \Big)^{2}\\\\\\a^{2} +2\cdot \underbrace{a\cdot \frac{1}{a} }_{1}+\frac{1}{a^{2} } =\frac{625}{144} \\\\\\a^{2} +2+\frac{1}{a^{2} } =\frac{625}{144} \\\\\\a^{2} +\frac{1}{a^{2} } =\frac{625}{144} -2\\\\\\\boxed{a^{2} +\frac{1}{a^{2} } =\frac{337}{144} }$

$\displaystyle\bf\\2)\\\\\Big(a+\frac{1}{a}\Big)^{3} = a^{3} +3\cdot a^{2} \cdot\frac{1}{a} +3\cdot a\cdot \frac{1}{a^{2} } +\frac{1}{a^{3} } \\\\\\\Big(a+\frac{1}{a} \Big)^{3} =a^{3} +3a+3\cdot \frac{1}{a} +\frac{1}{a^{3} } \\\\\\\Big(\underbrace{a+\frac{1}{a}}_{\dfrac{25}{12} } \Big)^{3} =a^{3} +3\cdot \Big(\underbrace{a+\frac{1}{a}}_{\dfrac{25}{12} } \Big)+\frac{1}{a^{3} } \\\\\\\frac{15625}{1728} =a^{3} +\frac{1}{a^{3} } +\frac{25}{4}$

$\displaystyle\bfd\\\boxed{a^{3} +\frac{1}{a^{3} } =\frac{4825}{1728} }\\\\\\3)\\\\a^{2} +\frac{1}{a^{2} } =\frac{337}{144} \\\\\\\Big(a^{2} +\frac{1}{a^{2} } \Big)^{2} =\Big(\frac{337}{144}\Big)^{2} \\\\\\a^{4} +2\cdot \underbrace{a^{2} \cdot \frac{1}{a^{2} } }_{1}+\frac{1}{a^{4} } =\frac{113569}{20736} \\\\\\a^{4} +\frac{1}{a^{4} } =\frac{113569}{20736} -2\\\\\\\boxed{{a^{4} +\frac{1}{a^4} } =\frac{72097}{20736} }$