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