Question

Здравствуйте, 43 или 44 на ваше усмотрение, с решением

Answer 1

$\displaystyle\bf\\43)\\\\a^{2} +\frac{1}{a^{2} } =a^{2} +\Big(\frac{1}{a} \Big)^{2} =\Big(a+\frac{1}{a} \Big)^{2} -2\cdot a\cdot \frac{1}{a} =\Big(a+\frac{1}{a} \Big)^{2} -2\\\\\\\Big(a+\frac{1}{a} \Big)^{2} -2=6\\\\\\\Big(a+\frac{1}{a} \Big)^{2} =8\\\\\\a+\frac{1}{a} =2\sqrt{2} \ \ tak \ \ kak \ \ a > 0\\\\\\a^{3} +\frac{1}{a^{3} } =a^{3} +\Big(\frac{1}{a} \Big)^{3} =\Big(a+\frac{1}{a} \Big)\Big(a^{2} -a\cdot\frac{1}{a} +\frac{1}{a^{2} } \Big)=\\\\\\=2\sqrt{2}\cdot(6-1 )=10\sqrt{2}$

$\displaystyle\bf\\44)\\\\a^{2} +\frac{1}{a^{2} } =a^{2} -\Big(\frac{1}{a} \Big)^{2} =\Big(a-\frac{1}{a} \Big)^{2} +2\cdot a\cdot \frac{1}{a} =\Big(a-\frac{1}{a} \Big)^{2} +2\\\\\\\Big(a-\frac{1}{a} \Big)^{2} +2=6\\\\\\\Big(a-\frac{1}{a} \Big)^{2} =4\\\\\\a-\frac{1}{a} =2 \ \ tak \ \ kak \ \ a > 1\\\\\\a^{3} -\frac{1}{a^{3} } =a^{3} -\Big(\frac{1}{a} \Big)^{3} =\Big(a-\frac{1}{a} \Big)\Big(a^{2} +a\cdot\frac{1}{a} +\frac{1}{a^{2} } \Big)=\\\\\\=2\cdot(6+1 )=14$