Question

efectuati :a) (4 radical din 27- 5 radical din 24 + 3 radical din 48): (-radical 3 )-( 12 radical din 2 +20): radical din 2

Answer 1

 

$\displaystyle\bf\\a)\\\\\Big(4\sqrt{27}-5\sqrt{24}+3\sqrt{48}\Big):\Big(-\sqrt{3}\Big)-\Big(12\sqrt{2}+20\Big):\Big(\sqrt{2}\Big)=\\\\\\=\frac{4\sqrt{27}-5\sqrt{24}+3\sqrt{48}}{-\sqrt{3}}-\frac{12\sqrt{2}+20}{\sqrt{2}}=\\\\\\=\frac{-4\sqrt{27}+5\sqrt{24}-3\sqrt{48}}{\sqrt{3}}-\frac{12\sqrt{2}+20}{\sqrt{2}}=\\\\\\=\frac{-4\cdot3\sqrt{3}+5\cdot2\sqrt{6}-3\cdot4\sqrt{3}}{\sqrt{3}}-\frac{12\sqrt{2}+10\cdot2}{\sqrt{2}}=$

.

$\displaystyle\bf\\=\frac{-4\cdot3\sqrt{3}+5\cdot2\sqrt{2}\cdot\sqrt{3}-3\cdot4\sqrt{3}}{\sqrt{3}}-\frac{12\sqrt{2}+10\cdot\sqrt{2}\cdot\sqrt{2}}{\sqrt{2}}=\\\\\\=\frac{\sqrt{3}\Big(-4\cdot3+5\cdot2\sqrt{2}-3\cdot4\Big)}{\sqrt{3}}-\frac{\sqrt{2}\Big(12+10\cdot\sqrt{2}\Big)}{\sqrt{2}}=\\\\\\=\Big(-4\cdot3+5\cdot2\sqrt{2}-3\cdot4\Big)-\Big(12+10\cdot\sqrt{2}\Big)=\\\\=\Big(-12+10\sqrt{2}-12\Big)-\Big(12+10\sqrt{2}\Big)=\\\\=\underline{10\sqrt{2}}-24-12-\underline{10\sqrt{2}}=-24-12=\boxed{\bf-36}$