Cześć!
Od a) do d)
[tex](x+1)^2=x^2+2\cdot x\cdot1+1^2=x^2+2x+1\\\\(x+2)^2=x^2+2\cdot x\cdot2+2^2=x^2+4x+4\\\\(x-3)^2=x^2-2\cdot x\cdot3+3^2=x^2-6x+9\\\\(x-5)^2=x^2-2\cdot x\cdot5+5^2=x^2-10x+25[/tex]
Od e) do h)
[tex](2x+1)^2=(2x)^2+2\cdot2x\cdot1+1^2=4x^2+4x+1\\\\(\frac{1}{2}x+2)^2=(\frac{1}{2}x)^2+2\cdot\frac{1}{2}x\cdot2+2^2=\frac{1}{4}x^2+2x+4\\\\(4x-1)^2=(4x)^2-2\cdot4x\cdot1+1^2=16x^2-8x+1\\\\(2x-\frac{1}{2})^2=(2x)^2-2\cdot2x\cdot\frac{1}{2}+(\frac{1}{2})^2=4x^2-2x+\frac{1}{4}[/tex]
