Question

5x-12+(3x-1)√3≤(√3+1)²x-√3(5x-√12)   
s={...-1,0,1}

Answer 1

Ai asa :
5x - 12 + 3x - 3$\sqrt{3}$ ≤ 3x + 2$\sqrt{3} x$ + x - 5$\sqrt{3} x$ - 6 <=> 8x -  2$\sqrt{3} x$ - x + 5$\sqrt{3} x$ ≤ 12 + 3$\sqrt{3}$
<=> 7x + 3$\sqrt{3} x$ ≤ 12 + 3$\sqrt{3}$ <=> x( 7 +  3$\sqrt{3}$ ) ≤ 12 + 3$\sqrt{3}$  <=>  x ≤ ( 12 + 3$\sqrt{3}$ ) / ( 7 +  3$\sqrt{3}$ ) => x ≤ (57 - 15$\sqrt{3} ) / 25$ => x ≤  1,24.

In multimea nr. intregi, ai solutii toate nr. intregi ≤ 1 => S = { _ _ _, -2, -1, 0 ,1 }

Bafta!