Question

Calculati:
a)$\sqrt{6}x( \sqrt{150}+ \sqrt{2 \frac{2}{3} } )-5x \sqrt{6}* \sqrt{2,(6)}=?$
b)$3(2 \sqrt{5}x-3 \sqrt{3}y-8 )-2( \sqrt{20}x- \sqrt{108}y-10 )$=?

Answer 1

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:

$\it \sqrt6 x \left(5\sqrt 6 + \dfrac{2\sqrt 6}{3}\right) - 5x\sqrt 6 \cdot \dfrac{2\sqrt 6}{3}$

$= \it 30x +\dfrac{2\cdot6}{3}x - \dfrac{10\cdot6}{3} x= 30x+4x-20x =14 x$

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]