Question

rationalizati numitorii. VA rog desfășurat. DAU COROANA
a)
$\frac{2}{ \sqrt{2} } =$
$\frac{3}{ \sqrt{3} } =$
$\frac{4}{ \sqrt{2} } =$
$\frac{2}{ \sqrt{8} } =$
$\frac{ \sqrt{6} }{ \sqrt{2} } =$
$\frac{ \sqrt{12} }{ \sqrt{3} } =$
b)
$\frac{ - \sqrt{24} }{2 \sqrt{3} } =$
$\frac{ - 6 \sqrt{7} }{ \sqrt{28} } =$
$\frac{15}{3 \sqrt{5} } =$
$\frac{ - 22}{2 \sqrt{11} } =$
$\frac{39}{2 \sqrt{13} } =$
$\frac{7}{ \sqrt{28} } =$
c)
$\frac{2 \sqrt{3} }{ \sqrt{2} } =$
$\frac{3 \sqrt{5} }{ \sqrt{15} } =$
$\frac{ - 4 \sqrt{6} }{ \sqrt{60} } =$
$\frac{ - \sqrt{50} }{ \sqrt{75} } =$
$\frac{20}{3 \sqrt{10} } =$
$\frac{12}{5 \sqrt{48} } =$
​

Answer 1

Răspuns:

=

\frac{3}{ \sqrt{3} } =

3

3

=

\frac{4}{ \sqrt{2} } =

2

4

=

\frac{2}{ \sqrt{8} } =

8

2

=

\frac{ \sqrt{6} }{ \sqrt{2} } =

2

6

=

\frac{ \sqrt{12} }{ \sqrt{3} } =

3

12

=

b)

\frac{ - \sqrt{24} }{2 \sqrt{3} } =

2

3

−

24

=

\frac{ - 6 \sqrt{7} }{ \sqrt{28} } =

28

−6

7

=

\frac{15}{3 \sqrt{5} } =

3

5

15

=

\frac{ - 22}{2 \sqrt{11} } =

2

11

−22

=

\frac{39}{2 \sqrt{13} } =

2

13

39

=

\frac{7}{ \sqrt{28} } =

28

7

=

c)

\frac{2 \sqrt{3} }{ \sqrt{2} } =

2

2

3

=

\frac{3 \sqrt{5} }{ \sqrt{15} } =

15

3

5

=

\frac{ - 4 \sqrt{6} }{ \sqrt{60} } =

60

−4

6

=

\frac{ - \sqrt{50} }{ \sqrt{75} } =

75

−

50

=

\frac{20}{3 \sqrt{10} } =

3

10

20

=

\frac{12}{5 \sqrt{48} } =

5

48

12

=