Question

Пусть a-(1/a)=2/3. Найти (a^12+1)/a^6.

Answer 1

Ответ:

(a^12+1)/a^6=a⁶+(1/a)⁶

a-(1/a)=2/3 возведем обе части в квадрат

a²+(1/a)²-2=4/9

a²+(1/a)²=22/9 третью стенень

(a⁴+(1/a)⁴+2)(a²+(1/a)²)=10648/729

а⁶+1/а²+2а²+а²+(1/а)⁶+2(1/а)²=10648/729

а⁶+(1/а)⁶+3(1/а²+a²)=10648/729

а⁶+(1/а)⁶=10648/729-22/3

а⁶+(1/а)⁶=(10648-5346)/729

а⁶+(1/а)⁶=5302/729

а⁶+(1/а)⁶=7 199/729

Перепроверяйте. Идея верная.

Объяснение:

Answer 2

$\displaystyle\bf\\a-\frac{1}{a} =\frac{2}{3} \\\\\\\Big(a-\frac{1}{a} \Big)^{2} =\Big(\frac{2}{3}\Big)^{2} \\\\\\a^{2} -2\cdot a\cdot \frac{1}{a} +\Big(\frac{1}{a}\Big)^{2} =\frac{4}{9} \\\\\\a^{2} -2+\frac{1}{a^{2} } =\frac{4}{9}\\\\\\a^{2} +\frac{1}{a^{2} } =\frac{22}{9} \\\\\\\Big(a^{2} +\frac{1}{a^{2} }\Big)^{3} =\Big(\frac{22}{9} \Big)^{3} \\\\\\a^{6} +3\cdot a^{4} \cdot\frac{1}{a^{2} } +3\cdot a^{2} \cdot\frac{1}{a^{4} } +\frac{1}{a^{6} } =\frac{10648}{729}$

$\displaystyle\bf\\a^{6} +3a^{2} +\frac{3}{a^{2} } +\frac{1}{a^{6} } =\frac{10648}{729} \\\\\\a^{6} +3\cdot\underbrace{\Big(a^{2} +\frac{1}{a^{2} } \Big)}_{\dfrac{22}{9} }+\frac{1}{a^{6} } =\frac{10648}{729} \\\\\\a^{6} +\frac{1}{a^{6} } =\frac{5302}{729} \\\\\\\frac{a^{12} +1}{a^{6} }=\frac{a^{12} }{a^{6} }+\frac{1}{a^{6} } =a^{6} +\frac{1}{a^{6} } =\frac{5302}{729} =7\frac{199}{729}$