Найдите значение выражения

1) $\frac{(\sqrt{7}+\sqrt{5})^2}{60+10\sqrt{35}} = \frac{7+2\sqrt{35}+5}{5(12+2\sqrt{35})} = \frac{12+2\sqrt{35}}{5(12+2\sqrt{35})} = \frac{1}{5} = 0,2$ 2) $\frac{(\sqrt{3}+\sqrt{7})^2}{40+8\sqrt{21}} = \frac{3+2\sqrt{21}+7}{4(10+2\sqrt{21})} = \frac{10+2\sqrt{21}}{4(10+2\sqrt{21})} = \frac{1}{4} = 0,25$ 3) $\frac{\sqrt{1,2} \cdot \sqrt{1,4}}{\sqrt{0,42}} = \frac{\sqrt{1,2 \cdot 1,4}}{\sqrt{0,42}} = \sqrt{\frac{1,68}{0,42}} = \sqrt{4} = 2$ 4) $\frac{\sqrt{1,5} \cdot \sqrt{3,3}}{\sqrt{0,55}} = \sqrt{\frac{1,5 \cdot 3,3}{0,55}} = \sqrt{\frac{4,95}{0,55}} = \sqrt{9} = 3$ 5) $(\sqrt{12}-\sqrt{75}) \cdot \sqrt{12} = \sqrt{12}\cdot\sqrt{12} - \sqrt{75}\cdot\sqrt{12} = 12 - \sqrt{900} = 12 - 30 = -18$ 6) $(\sqrt{63}-\sqrt{28}) \cdot \sqrt{7} = \sqrt{63}\cdot\sqrt{7} - \sqrt{28}\cdot\sqrt{7} = \sqrt{441} - \sqrt{196} = 21 - 14 = 7$ 7) $(\sqrt{32}-\sqrt{50}) \cdot \sqrt{8} = \sqrt{32}\cdot\sqrt{8} - \sqrt{50}\cdot\sqrt{8} = \sqrt{256} - \sqrt{400} = 16 - 20 = -4$ 8) $(\sqrt{75}-\sqrt{27}) \cdot \sqrt{3} = \sqrt{75}\cdot\sqrt{3} - \sqrt{27}\cdot\sqrt{3} = \sqrt{225} - \sqrt{81} = 15 - 9 = 6$