Краткий ответ:
- \(30\)
- \(3{,}4\)
- \(3\sqrt3-0{,}6\)
- \(400\)
Используем свойства: \[ \sqrt[n]{a}\cdot\sqrt[n]{b}=\sqrt[n]{ab},\qquad \frac{\sqrt[n]{a}}{\sqrt[n]{b}}=\sqrt[n]{\frac ab}. \]
1.
\[ \sqrt[4]{40\cdot250} +\sqrt[4]{128}\cdot\sqrt[4]{125}\cdot\sqrt[4]{10} \] \[ =\sqrt[4]{10000}+\sqrt[4]{128\cdot125\cdot10} =10+\sqrt[4]{160000}=10+20=\boxed{30}. \]
Проверка: \(10^4=10000,\;20^4=160000\).
2.
\[ \frac{\sqrt5\cdot\sqrt{12}}{\sqrt{15}}-1+ \sqrt[3]{0{,}064\cdot216} \] \[ =\sqrt{\frac{5\cdot12}{15}}-1+\sqrt[3]{13{,}824} =\sqrt4-1+2{,}4 =2-1+2{,}4=\boxed{3{,}4}. \]
3.
\[ \frac{\sqrt{90}\cdot\sqrt{240}}{\sqrt{800}} +\sqrt{0{,}1}\cdot\sqrt8\cdot\sqrt{0{,}2}-1 \] \[ =\sqrt{\frac{90\cdot240}{800}} +\sqrt{0{,}1\cdot8\cdot0{,}2}-1 \] \[ =\sqrt{27}+\sqrt{0{,}16}-1 =3\sqrt3+0{,}4-1 =\boxed{3\sqrt3-0{,}6}. \]
4.
\[ \sqrt2\cdot\sqrt5-\sqrt{10} +\frac{\sqrt[3]{10^3}}{\sqrt[3]{0{,}001}}\cdot\sqrt{16} \] \[ =\sqrt{10}-\sqrt{10} +\frac{10}{0{,}1}\cdot4 =0+100\cdot4=\boxed{400}. \]
