7 square root of 3 end root minus 4 square root of 6 end root plus square root of 48 end root minus square root of 54

1 answer:

5 0

$7\sqrt3-4\sqrt6+\sqrt{48}-\sqrt{54}\\\\=7\sqrt3-4\sqrt6+\sqrt{16\cdot3}-\sqrt{9\cdot6}\\\\=7\sqrt3-4\sqrt6+\sqrt{16}\cdot\sqrt3-\sqrt9\cdot\sqrt6\\\\=7\sqrt3-4\sqrt6+4\sqrt3-3\sqrt6\\\\=\boxed{11\sqrt3-7\sqrt6}$

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