Question

1)Fectueaza , folosind regula a(b+c)= ab+ ac .

(√2+√3) x √3. =

√5(√6 -√5) =

-√2(√8-√5 -√3) =

√7 (√2 -√3)

√6 (√5+√7) =

√3 (√3-√27 + √5)


2)Introdu Factorii sub radical :

2√5=

3√8=

11√2=

7√3=

√-2√5=

-5√7=

-10√2=

|-5| x √3 =

10√3=

1 pe 2 √30 =

- 4 pe 3 √6=

0,4√2=
0,(6)√6=

Answer 1

1.
(√2+√3) * √3. = √3*√3 + √3*√2 = 3 + √6

√5*(√6 -√5) = √5*√6 - √5*√5 = √30 - 5

-√2(√8-√5 -√3) = -√2*√8 + √2*√5 + √2*√3 = -√16 + √10 + √6 = √10 + √6 - 4

√7 (√2 -√3) = √7*√2 - √7*√3 = √14 - √21

√6 (√5+√7) = √6*√5 + √6*√7 = √30 + √35 

√3 (√3-√27 + √5) = √3*√3 - √3*√27 + √3*√5 = 3 - 9 - √15 = - 6 - √15

2.

2√5= √4*5 = √20

3√8= √9*8 = √72

11√2= √121*2 = √242

7√3=√49*3 = √147

-2√5=-√4*5 = -√20

-5√7=-√25*7 = -√175

-10√2= -√100*2 = -√200

10√3=√100*3 = √300

1 pe 2 √30 = √1/4*30