Odpowiedź:
[tex]a)\ \ (x+1)(x-1)+(x+2)(x-2)-(x+3)(x-3)=\\\\=x^2-1+x^2-4-(x^2-9)=2x^2-5-x^2+9=x^2+4\\\\\\dla\ \ x=\sqrt{3}\\\\x^2+4=(\sqrt{3})^2+4=3+4=7\\\\\\\\b)\\\\(1-2x)(1+2x)+(1-3x)(1+3x)-(1-4x)(4x+1)=\\\\=1-4x^2+1-9x^2-(1-4x)(1+4x)=2-13x^2-(1-16x^2)=\\\\=2-13x^2-1+16x^2=3x^2+1\\\\\\dla\ \ x=\sqrt{5}\\\\3x^2+1=3\cdot(\sqrt{5})^2+1=3\cdot5+1=15+1=16[/tex]
[tex]c)\ \ (2x-1)^2-(2x-1)(1+2x)-(2x+1)^2=\\\\=4x^2-4x+1-(2x-1)(2x+1)-(4x^2+4x+1)=\\\\=4x^2-4x+1-(4x^2-1)-4x^2-4x-1=-8x-4x^2+1=-4x^2-8x+1\\\\\\dla\ \ x=\sqrt{2}\\\\-4x^2-8x+1=-4\cdot(\sqrt{2})^2-8\sqrt{2}+1=-4\cdot2-8\sqrt{2}+1=-8-8\sqrt{2}+1=\\\\=-7-8\sqrt{2}[/tex]
