Question

porfavor me podrian ayudar.........................

Answer 1

Ejercicio 1:

$\sqrt[2]{\sqrt{\sqrt[3]{{a}^{36} \cdot{b}^{324}}}}$

$\sqrt[2]{ \sqrt[2]{\sqrt[3]{{a}^{36} \cdot{b}^{324}}}}$

$\sqrt[2 \times 2 \times 3]{{a}^{36} \cdot{b}^{324}}$

$\sqrt[12]{{a}^{36} \cdot{b}^{324}}$

$\sqrt[12]{{a}^{36}} \cdot\sqrt[12]{{b}^{324}}$

${a}^{ \frac{36}{12} } \cdot {b}^{ \frac{324}{12} }$

${a}^{3} \cdot {b}^{27}$

${a}^{3} \cdot {b}^{27} = {a}^{x} \cdot {b}^{y}$

Entonces:

• x=3
• y=27

Calcular:

$\sqrt[x]{ {y}^{2} } = \sqrt[3]{ {27}^{2} } = \sqrt[3]{{ ({3}^{3}) }^{2} } \\ = \sqrt[3]{ {( {3}^{2} )}^{3} } = \sqrt[3]{ {9 }^{3} } = 9$

Ejercicio 2

$(\sqrt[4]{\sqrt{x}}\cdot\sqrt[4]{\sqrt[4]{x}} \cdot \sqrt{ \sqrt{x} } )^{32}$

$(\sqrt[4]{ \sqrt[2]{x} }\cdot\sqrt[4]{\sqrt[4]{x}} \cdot \sqrt[2]{ \sqrt[2]{x} } )^{32}$

$(\sqrt[4 \times 2]{x}\cdot\sqrt[4 \times 4]{x} \cdot \sqrt[2 \times 2]{x} )^{32}$

$(\sqrt[8]{x}\cdot\sqrt[16]{x} \cdot \sqrt[4]{x} )^{32}$

$(\sqrt[8]{x})^{32} \cdot(\sqrt[16]{x})^{32} \cdot (\sqrt[4]{x} )^{32}$

$(x)^{ \frac{32}{8} } \cdot(x)^{ \frac{32}{16} } \cdot (x)^{ \frac{32}{4} }$

$(x)^{4} \cdot(x)^{2} \cdot (x)^{8}$

$x^{4 + 2 + 8}$

${x}^{14}$

Ejercicio 3

$\sqrt{x.\sqrt[5]{{x}^{2}.\sqrt[3]{{x}^{7}.\sqrt{{x}^{4}}}}}$

$\sqrt[2]{x.\sqrt[5]{{x}^{2}.\sqrt[3]{{x}^{7}. \sqrt[2]{{x}^{4}}}}}$

$\sqrt[2]{x.\sqrt[5]{{x}^{2}.\sqrt[3]{{x}^{7}. {x}^{ \frac{4}{2} } }}}$

$\sqrt[2]{x.\sqrt[5]{{x}^{2}.\sqrt[3]{{x}^{7}. {x}^{2} }}}$

$\sqrt[2]{x.\sqrt[5]{{x}^{2}.\sqrt[3]{{x}^{7 + 2} }}}$

$\sqrt[2]{x.\sqrt[5]{{x}^{2}.\sqrt[3]{{x}^{9}}}}$

$\sqrt[2]{x.\sqrt[5]{{x}^{2}. {x}^{ \frac{9}{3} } }}$

$\sqrt[2]{x.\sqrt[5]{{x}^{2}. {x}^{3} }}$

$\sqrt[2]{x.\sqrt[5]{{x}^{2 + 3}}}$

$\sqrt[2]{x.\sqrt[5]{{x}^{5}}}$

$\sqrt[2]{x. {x}^{ \frac{5}{5} } }$

$\sqrt[2]{x. {x}^{ 1} }$

$\sqrt[2]{ {x}^{1} . {x}^{ 1} }$

$\sqrt[2]{ {x}^{1 + 1} }$

$\sqrt[2]{ {x}^{2} }$

${x}^{ \frac{2}{2} }$

${x}^{1}$

El exponente final de "x" es 1