Bardzo proszę o pomoc. Dam najj za dobre rozwiązanie zad.1 Zapisz w postaci sumy algebraicznej. a)(x-5)² b)(2x+1)² c)(1/2x+2)² d)(2x-1/2)² zad.2 zapisz w postaci sumy algebraicznej a)(2x-1/4)² b)(√7+1)² c)(√5-3)²

[latex](x-5)^2=x^2-10x+25\(2x+1)^2=4x^2+4x+1\(frac{1}{2}x+2)^2=frac{1}{4}x^2+2x+4\(2x-frac{1}{2})^2=4x^2-2x+frac{1}{4}\(2x-frac{1}{4})^2=4x^2-x+frac{1}{16}\(sqrt7+1)^2=7+2sqrt7+1=8+2sqrt7\(sqrt5-3)^2=5-6sqrt5+9=14-6sqrt5[/latex]

[latex]zad 1\ a) (x-5)^{2}=x^{2}-2*x*5+5^{2}=x^{2}-10x+25\ \ b) (2x+1)^{2}=(2x)^{2}+2*2x*1+1^{2}=4x^{2}+4x+1\ \ c) (frac{1}{2}x+2)^{2}=(frac{1}{2}x)^{2}+2*frac{1}{2}x*2+2^{2}=frac{1}{4}x^{2}+2x+4\ \ d) (2x-frac{1}{2})^{2}=(2x)^{2}-2*2x*frac{1}{2}+(frac{1}{2})^{2}=4x^{2}-2x+frac{1}{4}\ ----------\[/latex] [latex]zad 2\ a) (2x-frac{1}{4})^{2}=(2x)^{2}-2*2x*frac{1}{4}+(frac{1}{4})^{2}=4x^{2}-x+frac{1}{16}\ \ b) (sqrt{7}+1)^{2}=sqrt{7}^{2}+2*sqrt{7}*1+1^{2}=7+2sqrt{7}+1=8+2sqrt{7}\ \ c) (sqrt{5}-3)^{2}=sqrt{5}^{2}-2*sqrt{5}*3+3^{2}=5-6sqrt{5}+9=14-6sqrt{5}[/latex]