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Błagam pomocy !!!! Poziom D



Błagam Pomocy Poziom D class=

Odpowiedź :

Amona
  • POZIOM D

a) [tex]4\sqrt{3} - \sqrt{27} - 2\sqrt{12} = 4\sqrt{3} - 3\sqrt{3} - 4\sqrt{3} = -3\sqrt{3}[/tex]

b) [tex]2\sqrt{125} - 3\sqrt{45} + 4\sqrt{5} = 10\sqrt{5} - 9\sqrt{5} + 4\sqrt{5} = 5\sqrt{5}[/tex]

c) [tex]\sqrt{200} - \sqrt{162} + \sqrt{2} = 10\sqrt{2} - 9\sqrt{2} + \sqrt{2} = 2\sqrt{2}[/tex]

d) [tex]\sqrt{63} + 3\sqrt{7} - \sqrt{28} = 3\sqrt{7} + 3\sqrt{7} - 2\sqrt{7} = 4\sqrt{7}[/tex]

e) [tex]5\sqrt{11} - \sqrt{44} +\sqrt{99} = 5\sqrt{11} - 2\sqrt{11} + 3\sqrt{11} = 6\sqrt{11}[/tex]

f) [tex]\sqrt{20} - \sqrt{45} + \sqrt{500} = 2\sqrt{5} - 3\sqrt{5} + 10\sqrt{5} = 9\sqrt{5}[/tex]

g) [tex]\sqrt{2} + \sqrt{8} - \sqrt{32} = \sqrt{2} + 2\sqrt{2} - 2^2\sqrt{2} = \sqrt{2} + 2\sqrt{2} - 4\sqrt{2} = -\sqrt{2}[/tex]

h) [tex]3\sqrt{3} + \sqrt{48} + \sqrt{75} = 3\sqrt{3} + 4\sqrt{3} + 5\sqrt{3} = 12\sqrt{3}[/tex]

i) [tex]2\sqrt{8} - \sqrt{18} + 5\sqrt{50} = 4\sqrt{2} - 3\sqrt{2} + 25\sqrt{2} = 26\sqrt{2}[/tex]

Magda

Odpowiedź:

[tex]Poziom\ \ D\\\\a)\ \ 4\sqrt{3}-\sqrt{27}-2\sqrt{12}=4\sqrt{3}-\sqrt{9\cdot3}-2\sqrt{4\cdot3}=4\sqrt{3}-3\sqrt{3}-2\cdot2\sqrt{3}=\\\\=\sqrt{3}-4\sqrt{3}=-3\sqrt{3}\\\\\\b)\ \ 2\sqrt{125}-3\sqrt{45}+4\sqrt{5}=2\sqrt{25\cdot5}-3\sqrt{9\cdot5}+4\sqrt{5}=2\cdot5\sqrt{5}-3\cdot3\sqrt{5}+4\sqrt{5}=\\\\=10\sqrt{5}-9\sqrt{5}+4\sqrt{5}=5\sqrt{5}[/tex]

[tex]c)\ \ \sqrt{200}-\sqrt{162}+\sqrt{2}=\sqrt{100\cdot2}-\sqrt{81\cdot2}+\sqrt{2}=10\sqrt{2}-9\sqrt{2}+\sqrt{2}=2\sqrt{2}\\\\\\d)\ \ \sqrt{63}+3\sqrt{7}-\sqrt{28}=\sqrt{9\cdot7}+3\sqrt{7}-\sqrt{4\cdot7}=3\sqrt{7}+3\sqrt{7}-2\sqrt{7}=4\sqrt{7}\\\\\\e)\ \ 5\sqrt{11}-\sqrt{44}+\sqrt{99}=5\sqrt{11}-\sqrt{4\cdot11}+\sqrt{9\cdot11}=5\sqrt{11}-2\sqrt{11}+3\sqrt{11}=6\sqrt{11}[/tex]

[tex]f)\ \ \sqrt{20}-\sqrt{45}+\sqrt{500}=\sqrt{4\cdot5}-\sqrt{9\cdot5}+\sqrt{100\cdot5}=2\sqrt{5}-3\sqrt{5}+10\sqrt{5}=\\\\=-\sqrt{5}+10\sqrt{5}=9\sqrt{5}\\\\\\g)\ \ \sqrt{2}+\sqrt{8}-\sqrt{32}=\sqrt{2}+\sqrt{4\cdot2}-\sqrt{16\cdot2}=\sqrt{2}+2\sqrt{2}-4\sqrt{2}=\\\\=3\sqrt{2}-4\sqrt{2}=-\sqrt{2}[/tex]

[tex]h)\ \ 3\sqrt{3}+\sqrt{48}+\sqrt{75}=3\sqrt{3}+\sqrt{16\cdot3}+\sqrt{25\cdot3}=3\sqrt{3}+4\sqrt{3}+5\sqrt{3}=12\sqrt{3}\\\\\\i)\ \ 2\sqrt{8}-\sqrt{18}+5\sqrt{50}=2\sqrt{4\cdot2}-\sqrt{9\cdot2}+5\sqrt{25\cdot2}=2\cdot2\sqrt{2}-3\sqrt{2}+5\cdot5\sqrt{2}=\\\\=4\sqrt{2}-3\sqrt{2}+25\sqrt{2}=26\sqrt{2}[/tex]

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