[tex]\displaystyle\bf\\1\frac{2}{5}x-1\frac{3}{4}=\frac{3}{5}x+0,\!5^2\\\\\frac{1\cdot5+2}{5}x-\frac{1\cdot4+3}{4}=\frac{3}{5}x+\Big(\frac{1}{2}\Big)^2\\\\\frac{7}{5}x-\frac{7}{4}=\frac{3}{5}x+\frac{1}{4}\\\\\frac{7}{5}x-\frac{3}{5}x=\frac{1}{4}+\frac{7}{4}\\\\\frac{7-3}{5}x=\frac{1+7}{4}\\\\\frac{4}{5}x=\frac{8}{4}\\\\\\x=\frac{~~\dfrac{8}{4}~~}{\dfrac{4}{5}}\\\\\\x=\dfrac{8}{4}\cdot\dfrac{5}{4}\\\\x=\frac{8\cdot5}{4\cdot4}\\\\x=\frac{40}{16}=\frac{5}{2}\\\\\boxed{\bf x=2\frac{1}{2}}[/tex]
