Cześć ;-)
Od a) do c)
[tex](x-2)(x+2)=x^2-2^2=\underline{x-4}\\\\(x-9)(x+9)=x^2-9^2=\underline{x^2-81}\\\\(2x-1)(2x+1)=(2x)^2-1^2=\underline{4x^2-1}[/tex]
Od d) do f)
[tex](6x+4x)(4x-6)=(4x)^2-6^2=\underline{16x^2-36}\\\\(3x-8y)(3x+8y)=(3x)^2-(8y)^2=\underline{9x^2-64y^2}\\\\(\frac{1}{2}x+y)(y-\frac{1}{2}x)=y^2-(\frac{1}{2}x)^2=\underline{y^2-\frac{1}{4}x^2}[/tex]
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