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Решите задачу:

$\sqrt{18}-\sqrt{32}+\sqrt{72}=\sqrt{9*2}-\sqrt{16*2}+\sqrt{36*2}=\\= \sqrt{3^2*2}-\sqrt{4^2*2}+\sqrt{6^2*2}=3\sqrt{2}-4\sqrt{2}+6\sqrt{2}=5\sqrt{2}\\\\\sqrt{72}+\sqrt{50}-\sqrt{162}=\sqrt{36*2}+\sqrt{25*2}-\sqrt{81*2}=\\=\sqrt{6^2*2}+\sqrt{5^2*2}-\sqrt{9^2*2}=6\sqrt{2}+5\sqrt{2}-9\sqrt{2}=2\sqrt{2}\\\\\sqrt{200}-\sqrt{8}-\sqrt{50}=\sqrt{100*2}-\sqrt{4*2}-\sqrt{25*2}=\\=\sqrt{10^2*2}-\sqrt{2^2*2}-\sqrt{5^2*2}=10\sqrt{2}-4\sqrt{2}-5\sqrt{2}=\sqrt{2}$

$\sqrt{128}+\sqrt{18}-\sqrt{98}=\sqrt{8^2*2}+\sqrt{3^2*2}-\sqrt{7^2*2}=\\=8\sqrt{2}+3\sqrt{2}-7\sqrt{2}=4\sqrt{2}\\\\10\sqrt{2}+\sqrt{8}+\sqrt{50}=10\sqrt{2}+\sqrt{2^2*2}+\sqrt{5^2*2}=\\=10\sqrt{2}+2\sqrt{2}+5\sqrt{2}=17\sqrt{2}$