$${\left(\frac{2^x+2^{-x}}{2}\right)}^2-{\left(\frac{2^x-2^{-x}}{2}\right)}^2+1$$

$$=\frac{{\left(2^x+{\displaystyle \frac{1}{2^x}}\right)}^2}{2^2}-\frac{{\left(2^x-{\displaystyle \frac{1}{2^x}}\right)}^2}{2^2}+1$$

$$=\frac{{\displaystyle {\left(\frac{2^x·2^x+1}{2^x}\right)}^2}}{{\displaystyle \frac{2^2}{1}}}-\frac{{\displaystyle {\left(\frac{2^x·2^x-1}{2^x}\right)}^2}}{{\displaystyle \frac{2^2}{1}}}+1$$

$$=\frac{{\displaystyle \frac{{\displaystyle {\left(2^{2x}+1\right)}^2}}{{\displaystyle {\left(2^x\right)}^2}}}}{{\displaystyle \frac{2^2}{1}}}-\frac{{\displaystyle \frac{{\displaystyle {\left(2^{2x}-1\right)}^2}}{{\displaystyle {\left(2^x\right)}^2}}}}{{\displaystyle \frac{2^2}{1}}}+1$$

$$=\frac{{\left(2^{2x}+1\right)}^2}{2^2·2^{2x}}-\frac{{\left(2^{2x}-1\right)}^2}{2^2·2^{2x}}+1$$

$$=\frac{{\left(2^{2x}+1\right)}^2}{2^{2x+2}}-\frac{{\left(2^{2x}-1\right)}^2}{2^{2x+2}}+1$$

$$=\frac{{\left(2^{2x}+1\right)}^2-{\left(2^{2x}-1\right)}^2}{2^{2x+2}}+1$$

$$=\frac{{\left(2^{2x}\right)}^2+2·2^{2x}·1+1-\left[{\left(2^{2x}\right)}^2-2^{2x}·2+1\right]}{2^{·+2}}+1$$

$$=\frac{2^{4x}+2·2^{2x}·1+1-\left[2^{4x}-2^{2x+1}+1\right]}{2^{·+2}}+1$$

$$=\frac{2^{4x}+2^{2x+1}+1-2^{4x}+2^{2x+1}-1}{2^{2x-2}}+1$$

$$=\frac{\cancel{2^{4x}}+2^{2x+1}\cancel{+1}\cancel{-2^{4x}}+2^{2x+1}\cancel{-}1}{2^{2x-2}}+1$$

$$=\frac{2·2^{2x+1}}{2^{2x+2}}+1$$

$$=\frac{2^{2x+1+1}}{2^{2x+2}}+1$$

$$=\frac{2^{2x+2}}{2^{2x+2}}+1$$

$$=1+1$$

$$=2$$