Question

Introduceți factorii sub radical. 49P.

Answer 1

3/7 √3 = √[(3/7)²×3 ] = √(27/49)

0,(3)√21 = √[(1/3)²×21] = √(21/9)

0,(1)√6 = √[(1/9)²×6 = √(6/81)

- 0,(6)√15 = -√[(2/3)² ×15 = -√(60/9)  

Answer 2

$\it a)\ \ \dfrac{3}{7}\sqrt3=\sqrt{\left(\dfrac{3}{7}\right)^2\cdot3}=\sqrt{\dfrac{9}{49}\cdot3}=\sqrt{\dfrac{27}{49}}\\ \\ \\ b)\ \ 0,(3)\sqrt{21}=\dfrac{\ 3^{(3}}{9}\sqrt{21}=\dfrac{1}{3}\sqrt{21} =\sqrt{\left(\dfrac{1}{3}\right)^2\cdot21}=\sqrt{\dfrac{1}{9}\cdot21}=\sqrt{\dfrac{7}{3}}\\ \\ \\ c)\ \ 0,(1)\sqrt6=\dfrac{1}{9}\sqrt6=\sqrt{\left(\dfrac{1}{9}\right)^2\cdot6}=\sqrt{\dfrac{1}{81}\cdot6}=\sqrt{\dfrac{\ 6^{(3}}{81}}=\sqrt{\dfrac{2}{27}}$

$\it d)\ \ -0,(6)\sqrt{15}=-\dfrac{\ 6^{(3}}{9}\sqrt{15}=-\dfrac{2}{3}\sqrt{15}=-\sqrt{\left(\dfrac{2}{3}\right)^2\cdot15}=-\sqrt{\dfrac{4}{9}\cdot15}=\\ \\ \\ =-\sqrt{\dfrac{\ 60^{(3}}{9}}=-\sqrt{\dfrac{20}{3}}$