Задача Задача №3762341
А
А
1.
- \(2^{1/2}\); 2) \(3^{-1/2}\); 3) \(2^{-3/2}\); 4) \(5^{2/3}\); 5) \(3^{3/4}\); 6) \(5^{-2/3}\).
2.
- \(\sqrt3\); 2) \(\frac1{2\sqrt2}\); 3) \(\sqrt[3]{25}\); 4) \(\sqrt[6]{3^5}\); 5) \(\sqrt[3]{(2/3)^{10}}\);
- \(\sqrt5\); 7) \(\frac1{\sqrt[4]2}\); 8) \(\sqrt[5]9\); 9) \(\sqrt{10}\); 10) \(\sqrt{\frac43}\).
3.
- \(2^{11/8}\);
- \(\left(\frac{2^2}{5^4}\right)^3=\frac{2^6}{5^{12}}\);
- \(3^{5/24}\);
- \(a^{17/6}\);
- \(a^{2/5}\);
- \(3^{-1/6}7^{1/3}\).
4.
- \(\frac12<2^{2/3}<2^{3/4}<2<2^{4/3}\);
- \(\left(\frac13\right)^2<9^{-1/3}<\left(\frac13\right)^{-1/2}<3^{3/4}<\left(\frac19\right)^{-3/2}\);
- \(\left(\frac9{25}\right)^2<\left(\frac34\right)^4<\left(-\frac34\right)^{-4}<2^4<(-3)^4\);
- \((-2)^3<\left(\frac23\right)^{-3}<3^3<2^6<\left(\frac1{27}\right)^{-2}\).
5.
- \(\left(\frac14\right)^{3/2}=\frac18>\frac19=\left(\frac1{27}\right)^{2/3}\);
- \(3^{1/3}>4^{1/4}>5^{1/5}\), так как \(3^4>4^3\) и \(4^5>5^4\);
- по неравенству Коши \(\sqrt{2003}+\sqrt{2005}<2\sqrt{2004}\);
- \(2^{1/100}>1\), поэтому \(2^{1/100}+10^{-1}>1\).
6.
- \(x=1\);
- \(x=1;\ -\frac34\);
- \(x=1\);
- \(x=-1;\ 2\);
- \(x=\frac13\);
- \(x=2\).
