Question

Помогите решить примеры

Answer 1

Ответ:

1.

а) $9.$

б) $8 + 2\sqrt{2}.$

2.

а) $x\sqrt[4]{x}.$

б) $\sqrt[9]{x^5}\ \ (x \geq 0).$

в) $\frac{1}{x\sqrt[12]{x}}\ \ (x \geq 0).$

г) $\frac{2}{50\sqrt[16]{x^7}}.$

3.

а) $64.$

б) $\frac{1}{64}.$

4.

$\sqrt{a}.$

5.

$\frac{2}{7}.$

Объяснение:

1.

а) $27^{\frac{2}{3}} = \left(\sqrt[3]{27} \right)^2 = 3^2 = 9.$

б) $16^{0,75} = 16^{\frac{3}{4}} = \left( \sqrt[4]{16} \right)^3 = 2^3 = 8.\\8 + 4\sqrt{\frac{1}{2}} = 8 + \sqrt{\frac{16}{2}} = 8 + \sqrt{8} = 8 + 2\sqrt{2}.$

2.

а) $x^{\frac{1}{2}} \cdot x^{\frac{3}{4}} = x^{\frac{1}{2} + \frac{3}{4}} = x^{\frac{5}{4}} = \sqrt[4]{x^5} = x\sqrt[4]{x}.$

б) $\left( x^{-\frac{5}{6}}\right)^{-\frac{2}{3}} = x^{-\frac{5}{6} \cdot (-\frac{2}{3})} = x^{\frac{10}{18}} = x^{\frac{5}{9}} = \sqrt[9]{x^5}\ \ (x \geq 0).$

в) $x^{-\frac{1}{3}}:x^{\frac{3}{4}} = x^{-\frac{1}{3} - \frac{3}{4}} = x^{-\frac{13}{12}} = \frac{1}{x^{\frac{13}{12}}} = \frac{1}{\sqrt[12]{x^{13}}} = \frac{1}{x\sqrt[12]{x}}. \ \ (x \geq 0).$

г) $\left(\frac{2}{50} x^{\frac{7}{8}}\right)^{-\frac{1}{2}} = \frac{2}{50}x^{-\frac{7}{16}} = \frac{2}{50x^{\frac{7}{16}}} = \frac{2}{50\sqrt[16]{x^7}}.$

3.

а) $x^{\frac{1}{3}} = 4 \Rightarrow \sqrt[3]{x} = 4 \Rightarrow \left( \sqrt[3]{x} \right)^3 = 4^3 \Rightarrow x = 64.$

б) $2x^{\frac{1}{6}} - 1^{\frac{1}{3}} = 0 \Rightarrow 2\sqrt[6]{x} = 1 \Rightarrow \sqrt[6]{x} = \frac{1}{2} \Rightarrow\\\Rightarrow \left( \sqrt[6]{x} \right)^6 = \left( \frac{1}{2} \right)^6 \Rightarrow x = \frac{1}{2^6} \Rightarrow x = \frac{1}{64}.$

4.

$\frac{(a + 3a^\frac{1}{2})}{a^\frac{1}{2}+3} = \frac{a^{\frac{1}{2}}(a^\frac{1}{2} + 3)}{(a^\frac{1}{2} + 3)} = a^\frac{1}{2} = \sqrt{a}.$

5.

$\left(y^\frac{1}{2} - 2\right)^{-1} - \left( y^\frac{1}{2} + 2\right)^{-1} = \frac{1}{\sqrt{y} - 2} - \frac{1}{\sqrt{y} + 2} = \frac{\sqrt{y} + 2 - (\sqrt{y} - 2)}{(\sqrt{y} - 2)(\sqrt{y} + 2)} = \frac{4}{y - 4};\\\\y = 18 \Rightarrow \frac{4}{18 - 4} = \frac{4}{14} = \frac{2}{7}.$