Bardzo łatwe zadanie, proszę o pomoc, wyznaczam najlepszą odpowiedź! :) Zapisz w postaci sumy algebraicznej: a) [latex] (s-3t)^{2} [/latex] b) [latex] (4p + 4q)^{2} [/latex] c) [latex] (-5k+4m)^{2} [/latex] d) [latex] (-7+2z)^{2} [/latex] e) [lat

[latex] (s-3t)^{2} =s^2-6st+9t^2\\ (4p + 4q)^{2}=16p^2+32pq+16q^2\\ (-5k+4m)^{2} =25k^2-40km+16m^2\\ (-7+2z)^{2} =49-28z+4z^2\\ (-5-6w)^{2} =25+60m+36w^2\\ (a sqrt{2}-b sqrt{5}) ^{2} =2a^2-2sqrt2*sqrt5*ab+5b^2=2a^2-2sqrt{10}*ab+5b^2[/latex]

[latex]a)\(s-3t)^2=s^2-2s*3t+9t^2=s^2-6t+9t^2\\b)\(4p+4q)^2=(4p)^2+2*4p*4q+(4q)^2=\\=16p^2+32pq+16q^2\c)\(-5k+4m)^2=(4m-5k)^2=\\=16m^2-40km+25k^2\\d)\(7+2z)^2=49+28z+4z^2\\e)\(-5-6w)^2=[-1(5+6w)]^2=\\=(-1)^2*(5+6w)^2=\\1*(25+60w+36w^2)=25+60w+36w^2\\f)\(asqrt2-bsqrt5)^2=(asqrt2)^2-2*asqrt2*bsqrt5+(bsqrt5)^2=\\=2a^2-2absqrt{10}+5b^2[/latex]