Obliczmy wartość danego wyrażenia. 

$\frac{1-{\left(\frac{3}{4}\right)}^{-\frac{1}{2}}}{1+{\left(\frac{3}{4}\right)}^{\frac{1}{2}}}-\frac{{\left(\frac{3}{4}\right)}^{\frac{1}{2}}+{\left(\frac{3}{4}\right)}^{-\frac{1}{2}}}{-\frac{1}{4}}=\frac{1-{\left(\frac{4}{3}\right)}^{\frac{1}{2}}}{1+\sqrt{\frac{3}{4}}}-\frac{\sqrt{\frac{3}{4}}+{\left(\frac{4}{3}\right)}^{\frac{1}{2}}}{-\frac{1}{4}}=\frac{1-\sqrt{\frac{4}{3}}}{1+\sqrt{\frac{3}{4}}}-\frac{\sqrt{\frac{3}{4}}+\sqrt{\frac{4}{3}}}{-\frac{1}{4}}=\frac{1-\frac{2}{\sqrt{3}}}{1+\frac{\sqrt{3}}{2}}-\frac{\frac{\sqrt{3}}{2}+\frac{2}{\sqrt{3}}}{-\frac{1}{4}}=\frac{1-\frac{2\sqrt{3}}{3}}{\frac{2}{2}+\frac{\sqrt{3}}{2}}-\frac{\frac{\sqrt{3}}{2}+\frac{2\sqrt{3}}{3}}{-\frac{1}{4}}=\frac{\frac{3}{3}-\frac{2\sqrt{3}}{3}}{\frac{2+\sqrt{3}}{2}}-\frac{\frac{3\sqrt{3}}{6}+\frac{4\sqrt{3}}{6}}{-\frac{1}{4}}=\frac{\frac{3-2\sqrt{3}}{3}}{\frac{2+\sqrt{3}}{2}}-\frac{\frac{3\sqrt{3}+4\sqrt{3}}{6}}{-\frac{1}{4}}=\frac{3-2\sqrt{3}}{3}\cdot \frac{2}{2+\sqrt{3}}-\frac{3\sqrt{3}+4\sqrt{3}}{{\cancel{6}}_3}\cdot \left(-\frac{{\cancel{4}}^2}{1}\right)=\frac{6-4\sqrt{3}}{6+3\sqrt{3}}+\frac{6\sqrt{3}+8\sqrt{3}}{3}=\frac{6-4\sqrt{3}}{6+3\sqrt{3}}+\frac{\left(6\sqrt{3}+8\sqrt{3}\right)\cdot \left(2+\sqrt{3}\right)}{6+3\sqrt{3}}=\frac{6-4\sqrt{3}}{6+3\sqrt{3}}+\frac{12\sqrt{3}+6\cdot 3+16\sqrt{3}+8\cdot 3}{6+3\sqrt{3}}=\frac{6-4\sqrt{3}}{6+3\sqrt{3}}+\frac{28\sqrt{3}+18+24}{6+3\sqrt{3}}=\frac{6-4\sqrt{3}+28\sqrt{3}+42}{6+3\sqrt{3}}=\frac{48+24\sqrt{3}}{6+3\sqrt{3}}=\frac{24\left(2+\sqrt{3}\right)}{3\left(2+\sqrt{3}\right)}=\frac{24}{3}=8$    

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Wiemy, że 4% pewnej liczby (oznaczmy ją jako x) wynosi 8, zatem:

$4\%\cdot x=8$  

$\frac{4}{100}\cdot x=8\mid :\frac{4}{100}$  

$x=8\cdot \frac{100}{4}$  

$x=8\cdot 25$  

$x=200$