Question

DAU 40 PUNCTE CLS 7
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$\sqrt{2} \times (3 \sqrt{3 } - 4 \sqrt{24} ) + (8 \sqrt{45} - 6 \sqrt{150} ) \div \sqrt{3}$
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Answer 1

$\sqrt2\cdot(3\sqrt3-4\sqrt{24})+(8\sqrt{45}-6\sqrt{150}):\srqt3=3\sqrt{2\cdot3}-4\sqrt{2\cdot24}+8\sqrt{45:3}-6\sqrt{150:3}=3\sqrt6-16\sqrt3+8\sqrt{15}-30\sqrt2$