Question

DAU CORONITA!! Introduceti factorii  sub radical: 5√6; 10√2,1; 1,6√10; 2\3√5; -3√7; (1-√2)√6; -x√5, x < 0; x²√x, x>0

Answer 1

5√5=√5²×5=√125
10√2,1=√10²×2,1=√210
1,6√10=1,6²×10=√25,6
2/3√5=√2/3²×5=√20/9
-3√7=√-3²×7=√-63
-x√5=√-x²×5=√-5x²

Answer 2

5√6=√(5²x6)=√150

10√2,1=√(10²x2,1)=√210

1,6√10=√(1,6²x10)=√25,6

2/3√5 unde este radicalul, la numitor?

-3√7=-(√(3²x7)=-√63

(1-√2)√6=√6-√12

-x√5=-√(5x²)

x²√x=√(x²)²x=√x⁵