Question

Ajutor chiar am nevoie fara a)​

Answer 1

a) $5\sqrt{12} + \sqrt{27} - 2\sqrt{75}  = \it 5 \cdot 2\sqrt{3} + 3\sqrt{3} - 2 \cdot 5\sqrt{3} = 10\sqrt{3} + 3\sqrt{3} - 10\sqrt{3} =3\sqrt{3}$

b) $3\sqrt{20} - 4\sqrt{45} + \sqrt{125} = \it 3 \cdot 2\sqrt{5} - 4 \cdot 3\sqrt{5} + 5\sqrt{5} = 6\sqrt{5} - 12\sqrt{5} + 5\sqrt{5} = -\sqrt{5}$

c) $\sqrt{54} - 3\sqrt{96} + 4\sqrt{150} = \it 3\sqrt{6} - 3 \cdot 4\sqrt{6} + 4 \cdot 5\sqrt{6} = 3\sqrt{6} - 12\sqrt{6} + 20\sqrt{6} = - 9\sqrt{6} + 20\sqrt{6} = 11\sqrt{6}$

d) $5\sqrt{28} + 2\sqrt{63} - 7\sqrt{175} = \it 5 \cdot 2\sqrt{7} + 2 \cdot 3\sqrt{7} - 7 \cdot 5\sqrt{7} = 10\sqrt{7} + 6\sqrt{7} - 35\sqrt{7} = 16\sqrt{7} - 35\sqrt{7} = -19\sqrt{7}$

e) $-7\sqrt{48} + 2\sqrt{108} + \sqrt{75} = \it -7 \cdot 4\sqrt{3} + 2 \cdot 6\sqrt{3} + 5\sqrt{3} = -28\sqrt{3} + 12\sqrt{3} + 5\sqrt{3} = -16\sqrt{3} + 5\sqrt{3} = -11\sqrt{3}$