Sa procedam metodic ( si eficient) :
[tex]\sqrt{150} =\sqrt{25\cdot6} =5\sqrt6
\\\;\\
\sqrt{2 \dfrac{2}{3}} = \sqrt{\dfrac{^{3)}8}{\ 3}} = \sqrt{\dfrac{24}{9}} = \dfrac{\sqrt{24}}{3}=\dfrac{\sqrt{4\cdot6}}{3}= \dfrac{2\sqrt6}{3}
\\\;\\
\sqrt{2,(6)} =\sqrt{2\dfrac{6}{9}} = \sqrt{\dfrac{24}{9}} = \dfrac{2\sqrt6}{3}
\\\;\\
.[/tex]
Acum, expresia de calculat devine:
[tex]\it \sqrt6 x \left(5\sqrt 6 + \dfrac{2\sqrt 6}{3}\right) - 5x\sqrt 6 \cdot \dfrac{2\sqrt 6}{3}[/tex]
[tex]= \it 30x +\dfrac{2\cdot6}{3}x - \dfrac{10\cdot6}{3} x= 30x+4x-20x =14 x[/tex]
b)
[tex]\sqrt{20} =\sqrt{4\cdot5} =2\sqrt5
\\\;\\
\sqrt{108}= \sqrt{36\cdot3}=6\sqrt3
.[/tex]
Expresia devine:
[tex]3(2\sqrt5-3\sqrt3 y-8)-2(2\sqrt5 x -6\sqrt3y-10) =
\\\;\\
=6\sqrt5x-9\sqrt3y-24 -4\sqrt5x+12\sqrt3 y+20 = 2\sqrt5 x+3\sqrt3 y-4
\\\;\\
.[/tex]
