Odpowiedź:
[tex]\sqrt{3}+\sqrt{12}=\sqrt{3}+\sqrt{4\cdot3}=\sqrt{3}+2\sqrt{3}=3\sqrt{3}\\\\\sqrt{54}-\sqrt{6}=\sqrt{9\cdot6}-\sqrt{6}=3\sqrt{6}-\sqrt{6}=2\sqrt{6}\\\\\sqrt{50}-\sqrt{2}=\sqrt{25\cdot2}-\sqrt{2}=5\sqrt{2}-\sqrt{2}=4\sqrt{2}\\\\\sqrt{10}+\sqrt{90}=\sqrt{10}+\sqrt{9\cdot10}=\sqrt{10}+3\sqrt{10}=4\sqrt{10}\\\\\sqrt{28}-\sqrt{7}=\sqrt{4\cdot7}-\sqrt{7}=2\sqrt{7}-\sqrt{7}=\sqrt{7}\\\\\sqrt{45}+\sqrt{80}=\sqrt{9\cdot5}+\sqrt{16\cdot5}=3\sqrt{5}+4\sqrt{5}=7\sqrt{5}[/tex]
