Question

ajuta ti ma va rog la ex A6, DAU COROANA

Answer 1

Explicație pas cu pas:

a)

$\sqrt{3 {x}^{6} {y}^{5}} = \sqrt{3 {( {x}^{3} )}^{2} {( {y}^{2} )}^{2}y} = \sqrt{ {({x}^{3} {y}^{2})}^{2}3y } = {x}^{3} {y}^{2} \sqrt{3y}$

b)

$\sqrt[4]{5 {x}^{4} {y}^{5} } = \sqrt[4]{5 {x}^{4} {y}^{4} y} = \sqrt[4]{ {(xy)}^{4} 5y} = xy \sqrt[4]{5y}$

c)

$\sqrt[3]{27{x}^{3} {y}^{6} {z}^{4} } = \sqrt[3]{ {3}^{3} {x}^{3} {( {y}^{2} )}^{3} {z}^{3} z} = \sqrt[3]{ {(3x {y}^{2}z)}^{3}z } = 3x {y}^{2}z \sqrt[3]{z}$

d)

$\sqrt{ \frac{1}{49} {x}^{3} {y}^{6} } = \sqrt{ \frac{1}{ {7}^{2} } {x}^{2}x{( {y}^{3} )}^{2} } = \frac{ {x}^{2} {y}^{3} }{7} \\\sqrt{x}$

e)

$\sqrt[3]{ \sqrt{ {x}^{7} {y}^{12} } } = \sqrt[6]{{x}^{7} {y}^{12}} = \sqrt{ {(x {y}^{2} )}^{6} x} = x {y}^{2} \sqrt{x}$