Question

(radical 3+1)^2 x-(3x-1)radical 3=5x-12+radical 3(5x-radical 12)

Answer 1

(rad 3+1)^2 x-(3x-1)rad 3 =5x-12+rad 3(5x-rad 12)

rad=radical

(3+1)x-(3x-1)x 2rad3=5x-12rad3×(5x×2rad3)=

4x-(6rad3 x -2rad3)=5x-12rad3×(10rad3 x)=4x-6rad3+2rad3=5x-12rad3 × 10rad3 x=

4x-6rad3+2rad3=5x-12×3×10x=4x-6rad3 x + 2rad3=5x-360x=4x-6rad3+2rad3=355x=4x-6rad3 x +355x= -2rad3=359x-6rad3 x =-2rad3=(359-6rad3)x=-2rad3=x-2rad3/259-6rad3=x-2rad3×(359+6rad3)=359^2-108=x=718rad3+36/359^2-108=x=0,00993696( forma alternativa

Answer 2

$( \sqrt{3 + 1} ) {}^{2} \times x - (3x - 1) \sqrt{3} = 5x - 12 + \sqrt{3} (5x - \sqrt{12} )$

$(3 + 1)x - (3x - 1) \sqrt{3} = 5x - 12 + \sqrt{3} (5x - \sqrt{12}$

$(3 + 1)x - (3 \sqrt{3} x - \sqrt{3} ) = 5x - 12 + \sqrt{3} (5x - 2 \sqrt{3}$

$4x - (3 \sqrt{3} - \sqrt{3} ) = 5x - 12 + \sqrt{3} (5x - 2 \sqrt{3}$

$4x - 3 \sqrt{3} + \sqrt{3} = 5x - 12 + 5 \sqrt{3} x - 6$

$4x - 3 \sqrt{3} x + \sqrt{3} = 5x - 18 + 5 \sqrt{3} x$

$4x - 3 \sqrt{3} x - 5x - 5 \sqrt{3} x + \sqrt{3} = - 18$

$4x - 3 \sqrt{3} x - 5x - 5 \sqrt{3} x = - 18 - \sqrt{3}$

$- x - 3 \sqrt{3} x - 5 \sqrt{3} x = - 18 - \sqrt{3}$

$- x - 8 \sqrt{3} x = - 18 - \sqrt{3}$

$( - 1 - 8 \sqrt{3} )x = - 18 - \sqrt{3}$

$x = \frac{ - 18 - \sqrt{3} }{ - 1 - 8 \sqrt{3} }$

$x = \frac{( - 18 - \sqrt{3}) ( - 1 + 8 \sqrt{3} ) }{ - 191}$

$x = \frac{18 - 144 \sqrt{3} + \sqrt{3 } - 24 }{ - 191}$

$x = - \frac{ - 6 - 143 \sqrt{3} }{ 191}$

$x = \frac{6 + 143 \sqrt{3} }{191}$