Question

√48 - (16/√32-18/√243)*√6

Answer 1

[tex]\it \sqrt{48} = \sqrt{16\cdot3} = 4\sqrt3 \\\;\\ \sqrt{32} = \sqrt{16\cdot2} = 4\sqrt2 \\\;\\ \sqrt{243} = \sqrt{81\cdot3} = 9\sqrt3 \\\;\\ \sqrt6=\sqrt2\cdot\sqrt3 [/tex]

Acum, expresia devine :

[tex]\it 4\sqrt3-\left(\dfrac{16}{4\sqrt2}-\dfrac{18}{9\sqrt3}\right)\cdot\sqrt2\cdot\sqrt3 =4\sqrt3-\left(\dfrac{4}{\sqrt2}-\dfrac{2}{\sqrt3}\right)\sqrt2\cdot\sqrt3 = \\\;\\ \\\;\\ =4\sqrt3-\left(\dfrac{4}{\sqrt2}\cdot\sqrt2\cdot\sqrt3-\dfrac{2}{\sqrt3}\cdot\sqrt2\cdot\sqrt3\right) = 4\sqrt3-4\sqrt3+2\sqrt2= \\\;\\ \\\;\\ =2\sqrt2.[/tex]