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Решите задачу:

$\frac{\sqrt{3\frac{6}{7}}-\sqrt{1\frac{5}7}}{\sqrt{\frac{3}{28}}}=\frac{\sqrt{\frac{27}{7}}}{\sqrt{\frac{3}{28}}}-\frac{\sqrt{\frac{12}{7}}}{\sqrt{\frac{3}{28}}}=\\\\=\sqrt{\frac{\frac{27}{7}}{\frac{3}{28}}}-\sqrt{\frac{\frac{12}{7}}{\frac{3}{28}}}=\sqrt{\frac{28*27}{3*7}}-\sqrt{\frac{28*12}{3*7}}=\\\\=\sqrt{9*4}-\sqrt{4*4}=\sqrt{36}-\sqrt{16}=6-4=2.$