Question

Efectua las siguientes redicaciones:​

Answer 1

Hola!

Explicación paso a paso:

f)

$\sqrt{40xy + \sqrt{80 {x}^{2} {y}^{2} + \sqrt{ {x}^{4} {y}^{4} }} } \\ = \sqrt{40xy + \sqrt{80 {x}^{2} {y}^{2} + \sqrt{ {( {x}^{2} {y}^{2} )}^{2} } } } \\ = \sqrt{40xy + \sqrt{80 {x}^{2} {y}^{2} + {x}^{2} {y}^{2} } } \\ = \sqrt{40xy + \sqrt{81 {x}^{2} {y}^{2} } } \\ = \sqrt{40xy + \sqrt{ {(9xy)}^{2} } } \\ = \sqrt{40xy + 9xy} \\ = \sqrt{49xy} \\ = \sqrt{49} \sqrt{xy} \\ = 7 \sqrt{xy}$

e)

$\sqrt{44 + \sqrt{22 + \sqrt{9} } } \\ = \sqrt{44 + \sqrt{22 + 3} } \\ = \sqrt{44 + \sqrt{25} } \\ = \sqrt{44 + 5} \\ = \sqrt{49} \\ = 7$

d)

$\sqrt{32a + \sqrt{8 {a}^{2} + \sqrt{64 {a}^{4} } } } \\ = \sqrt{32a + \sqrt{8 {a}^{2} + \sqrt{ {(8 {a}^{2}) }^{2} } } } \\ = \sqrt{32a + \sqrt{8 {a}^{2} + 8 {a}^{2} } } \\ = \sqrt{32a + \sqrt{16 {a}^{2} } } \\ = \sqrt{32a + \sqrt{ {(4a)}^{2} } } \\ = \sqrt{32a + 4a} \\ = \sqrt{36a} \\ = \sqrt{36} \sqrt{a} \\ = 6 \sqrt{a}$

c)

$\sqrt{3x \sqrt[3]{3 \sqrt{3} } } \\ = \sqrt{3x \sqrt[3]{3 \times {3}^{\frac{1}{2} }} } \\ = \sqrt{3x \sqrt[3]{ {3}^{1 + \frac{1}{2} } } } \\ = \sqrt{3x \sqrt[3]{ {3}^{ \frac{3}{2} } } } \\ = \sqrt{3 \times {3}^{ \frac{1}{2} } } \\ = \sqrt{ {3}^{1 + \frac{1}{2} } } \\ = \sqrt{ {3}^{ \frac{3}{2} } } \\ = {3}^{ \frac{3}{4} } \\ = \sqrt[4]{ {3}^{3} } \\ = \sqrt[4]{27}$

b)

$\sqrt{ \sqrt[5]{225 {x}^{6} {y}^{4} } } \\ = \sqrt{ \sqrt[5]{ {(15 {x}^{3} {y}^{2}) }^{2} } } \\ = \sqrt[10]{ {(15 {x}^{3} {y}^{2}) }^{2} } \\ = \sqrt[5]{15 {x}^{3} {y}^{2} }$

a)

$\sqrt{ \sqrt[4]{256 {x}^{8} } } \\ = \sqrt[8]{ {2}^{8} {x}^{8} } \\ = \sqrt[8]{ {(2x)}^{8} } \\ = 2x$