Tъждествата сбор и разлика на квадрат

[Гост]

Благодаря!!!

[ammornil]

[tex]\left(M-\frac{1}{2}xy\right)^{2}=9z^{4}-Nz^{2}+\frac{1}{4}x^{2}y^{2} \\ M^{2}-2\cdot{}M\cdot{}\frac{1}{2}xy+\left(\frac{1}{2}xy \right)^{2}=9z^{4}-Nz^{2}+\frac{1}{4}x^{2}y^{2} \\ \begin{array}{|l} M^{2}=9z^{4} \\ Nz^{2}=M\cdot{xy} \end{array} \Leftrightarrow \begin{array}{|l} M=3z^{2} \\ nz^{2}=3z^{2}\cdot{xy} \end{array} \Leftrightarrow \begin{array}{|l} M=3z^{2} \\ N=3xy \end{array}[/tex]

[ammornil]

[tex]A=\frac{(2^{-3}-x)^{2}+(2^{-3}+x)^{2}}{x^{2}\left(\frac{1}{4^{3}}+x^{2} \right)}=?, \quad x=\frac{1}{3} \\ \quad (2^{-3}-x)^{2}+(2^{-3}+x)^{2}=(2^{-3})^{2}\cancel{-2\cdot{}2^{-3}\cdot{x}}+x^{2}+(2^{-3})^{2}\cancel{+2\cdot{2^{-3}}\cdot{x}}+x^{2}=2\cdot{2^{-6}}+2x^{2}=2^{-5}+2x^{2}=\frac{1}{32}+2\cdot{}\frac{1}{9}=\frac{1}{32}+\frac{2}{9}=\frac{73}{9\cdot{32}} \\ \quad x^{2}\left(\frac{1}{4^{3}}+x^{2} \right)=\frac{1}{9}\cdot{\left(\frac{1}{64}+\frac{1}{9} \right)}=\frac{1}{9}\cdot{}\frac{73}{9\cdot{64}}=\frac{73}{9^{2}\cdot{64}} \\ A=\frac{\frac{73}{9\cdot{32}}}{\frac{73}{9^{2}\cdot{64}}}=\frac{73\cdot{}9^{2}\cdot{}64}{73\cdot{}9\cdot{}32}=9\cdot{2}=18[/tex]