Welke gelijkheid is correct ?
A.  \(\sqrt 1 + \sqrt 1 + \sqrt 1 = \sqrt 1 \cdot \sqrt 1 \cdot \sqrt 1\)
B.  \(\sqrt 3 + \sqrt 3 + \sqrt 3 = \sqrt 3 \cdot \sqrt 3 \cdot \sqrt 3\)
C.  \(\sqrt 5 + \sqrt 5 + \sqrt 5 = \sqrt 5 \cdot \sqrt 5 \cdot \sqrt 5\)
D.  \(\sqrt 7 + \sqrt 7 + \sqrt 7 = \sqrt 7 \cdot \sqrt 7 \cdot \sqrt 7\)
E.  \(\sqrt 9 + \sqrt 9 + \sqrt 9 = \sqrt 9 \cdot \sqrt 9 \cdot \sqrt 9\)
F.  geen van de vorige
A   B   C   D   E   F

[ 3-3604 - op net sinds 10.7.15-(E)-20.10.2026 ]

Translation in   E N G L I S H

Which equality is correct ?
A.  \(\sqrt 1 + \sqrt 1 + \sqrt 1 = \sqrt 1 \cdot \sqrt 1 \cdot \sqrt 1\)
B.  \(\sqrt 3 + \sqrt 3 + \sqrt 3 = \sqrt 3 \cdot \sqrt 3 \cdot \sqrt 3\)
C.  \(\sqrt 5 + \sqrt 5 + \sqrt 5 = \sqrt 5 \cdot \sqrt 5 \cdot \sqrt 5\)
D.  \(\sqrt 7 + \sqrt 7 + \sqrt 7 = \sqrt 7 \cdot \sqrt 7 \cdot \sqrt 7\)
E.  \(\sqrt 9 + \sqrt 9 + \sqrt 9 = \sqrt 9 \cdot \sqrt 9 \cdot \sqrt 9\)
F.  none of thesee

Oplossing - Solution

√3 + √3 + √3 = 3.√3
√3 . √3 . √3 = (√3 . √3) . √3 = 3 . √3
gricha