Odpowiedź :
Odpowiedź:
A) [tex]\sqrt{32} -\sqrt{18} -\sqrt{8} =4\sqrt{2} -3\sqrt{2} -2\sqrt{2} =4\sqrt{2} -5\sqrt{2} =-\sqrt{2}[/tex]
B)[tex]\sqrt{20} +\sqrt{45} -\sqrt{125} =2\sqrt{5} +3\sqrt{5} -5\sqrt{5} =0[/tex]
chyba w przykładzie B jest błąd w zapisie,ale gdyby zapis był prawidłowy to:
[tex]\sqrt{20} +\sqrt{45} -125=2\sqrt{5} +3\sqrt{5} -125=5\sqrt{5} -125[/tex]
C)[tex]\sqrt{12} +\sqrt{27} -\sqrt{75}=2 \sqrt{3} +3\sqrt{3} -5\sqrt{3} =0[/tex]
D)[tex]\sqrt{160} -2\sqrt{90} +3\sqrt{40} =4\sqrt{10} -2*3\sqrt{10} +3*2\sqrt{10} =4\sqrt{10} -6\sqrt{10} +6\sqrt{10} =4\sqrt{10}[/tex]
Szczegółowe wyjaśnienie:
[tex]A.\\\sqrt{32} - \sqrt{18}-\sqrt{8} =\sqrt{16\cdot2} - \sqrt{9\cdot2} - \sqrt{4\cdot2} = \sqrt{16}\cdot\sqrt{2}-\sqrt{9}\cdot\sqrt{2}-\sqrt{4}\cdot\sqrt{2} = \\\\=4\sqrt{2}-3\sqrt{2}-2\sqrt{2} = 4\sqrt{2}-5\sqrt{2} = -\sqrt{2}[/tex]
[tex]B.\\\sqrt{20}+\sqrt{45} - \sqrt{125} =\sqrt{4\cdot5} + \sqrt{9\cdot5}-\sqrt{25\cdot5} = \sqrt{4}\cdot\sqrt{5}+\sqrt{9}\cdot\sqrt{5}-\sqrt{25}\cdot\sqrt{5} =\\\\=2\sqrt{5}+3\sqrt{5}-5\sqrt{5} = 5\sqrt{5}-5\sqrt{5} = 0[/tex]
[tex]C.\\\sqrt{12}+\sqrt{27}-\sqrt{75} = \sqrt{4\cdot3} + \sqrt{9\cdot3} -\sqrt{25\cdot3} = \sqrt{4}\cdot\sqrt{3}+\sqrt{9}\cdot\sqrt{3} -\sqrt{25}\cdot\sqrt{3} =\\\\=2\sqrt{3}+3\sqrt{3}-5\sqrt{3} = 5\sqrt{3}-5\sqrt{3} = 0[/tex]
[tex]D.\\\sqrt{160} - 2\sqrt{90} + 3\sqrt{40} = \sqrt{16\cdot10} - 2\cdot\sqrt{9\cdot10} +3\cdot\sqrt{4\cdot10} =\\\\= \sqrt{16}\cdot\sqrt{10} -2\cdot\sqrt{9}\cdot\sqrt{10}+3\cdot\sqrt{4}\cdot\sqrt{10} = 4\sqrt{10}-2\cdot3\sqrt{10} +3\cdot2\sqrt{10}=\\\\=4\sqrt{10}-6\sqrt{10}+6\sqrt{10} = 4\sqrt{10}[/tex]