Question

16 supra 2-√3 + 12 supra √3+1 - 6 supra 2√3-3 - 8 supra 3√3+5

Answer 1

16(2+√3)/1 +12(√3 -1)/2 -6(2√3 +3)/3 - 8(3√3 -5)/ 2 =
=  16(2+√3) + (12√3 -12 -24√3 + 40)/2 - 2(2√3 + 3) 
=  32 +16√3 - 6√3 + 14 - 4√3 - 6
= 40 + 6√3 

Answer 2

Daca o fractie are la numitor a-b, o amplificam cu a+b si invers
$=\frac{16(2+\sqrt{3})}{(2-\sqrt{3})(2+\sqrt{3})}+\frac{12(\sqrt{3}-1)}{(\sqrt{3}+1)(\sqrt{3}-1)}-\frac{6(2\sqrt{3}+3)}{(2\sqrt{3}-3)(2\sqrt{3}+3)}-\frac{8(3\sqrt{3}-5)}{(3\sqrt{3}+5)(3\sqrt{3}-5)} \\ \\ =\frac{16(2+\sqrt{3})}{2^2-(\sqrt{3})^2}+\frac{12(\sqrt{3}-1)}{(\sqrt{3})^2-1^2}-\frac{6(2\sqrt{3}+3)}{(2\sqrt{3})^2-3^2}-\frac{8(3\sqrt{3}-5)}{(3\sqrt{3})^2-5^2} \\ \\ =\frac{16(2+\sqrt{3})}{4-3}+\frac{12(\sqrt{3}-1)}{3-1}-\frac{6(2\sqrt{3}+3)}{12-9}-\frac{8(3\sqrt{3}-5)}{27-25} \\ \\ =\frac{16(2+\sqrt{3})}{1}+\frac{12(\sqrt{3}-1)}{2}-\frac{6(2\sqrt{3}+3)}{3}-\frac{8(3\sqrt{3}-5)}{2} \\ \\ =16(2+\sqrt{3})+6(\sqrt{3}-1)-2(2\sqrt{3}+3)-4(3\sqrt{3}-5) \\ \\ =32+16\sqrt{3}+6\sqrt{3}-6-(4\sqrt{3}+6)-(12\sqrt{3}-20) \\ \\ 26+22\sqrt{3}-4\sqrt{3}-6-12\sqrt{3}+20 \\ \\ =26+18\sqrt{3}-6-12\sqrt{3}+20 \\ \\ =20+18\sqrt{3}-12\sqrt{3}+20 \\ \\ =20+6\sqrt{3}+20 \\ \\ =40+6\sqrt{3}$