a)
(\sqrt2-x)^2-(2x^3+1)(2x^3-1)= …x=\sqrt2
=2-2\sqrt2x+x^2-[(2x^3)^2-1]=
=2-2\sqrt2x+x^2-4x^6+1=
=-4x^6+x^2-2\sqrt2x+3=
podstawiam x
{=-4*(\sqrt2)^6+(\sqrt2)^2-2\sqrt2*\sqrt2+3=-4*(\sqrt2)^2)^3+2-2*2+3=-4*2^3+2-4+3=-31}
b)
(x+2)^2-(3x+2)(2-3x)-(\sqrt3+x)^2= …x=\sqrt3
=x^2+4x+4-(2^2-9x^2)-(3+2\sqrt3x+x^2)=
=x^2+4x+4-4+9x^2-3-2\sqrt3x-x^2=
=9x^2-2\sqrt3x+4x-3=
podstawiam x
=9*(\sqrt3)^2-2\sqrt3*\sqrt3+4\sqrt3-3=9*3-2*3+4\sqrt3-3=18+4\sqrt3
c)
(x-3)^2+\sqrt5(2x+\sqrt5)(\sqrt5-2x)-(1-x)^2=… x=-\sqrt5
=x^2-6x+9+\sqrt5[(\sqrt5)^2-(2x)^2]-(1-2x+x^2=
=x^2-6x+9+\sqrt5(5-4x^2)-1+2x-x^2=
=-4\sqrt5x^2-4x+5\sqrt5+8=
podstawiam x
{=-4\sqrt5*(-\sqrt5)^2+4\sqrt5+5\sqrt5+8=-4\sqrt5*5+9\sqrt5+8=-20\sqrt5+9\sqrt5+8=8-11\sqrt5}