\left \{ {{(\frac{1}{2}x-1)^2 - (\frac{3}{4}x + 2)^2 = 0} \atop {x \geq -\frac{8}{3} }} \right. => \left \{ {{(\frac{1}{2}x-1 -(\frac{3}{4}x + 2))(\frac{1}{2}x-1 +\frac{3}{4}x + 2)=0 } \atop {x \geq -\frac{8}{3}}} \right. =>\\ \left \{ {{(-\frac{1}{4}x-3)(\frac{5}{4} x+1)= 0} \atop {x\geq-\frac{8}{3} }} \right. =>" alt="|\frac{1}{2}x-1| = \frac{3}{4}x + 2\\\left \{ {{(\frac{1}{2}x-1)^2 = (\frac{3}{4}x + 2)^2} \atop {\frac{3}{4}x + 2 \geq 0 }} \right. => \left \{ {{(\frac{1}{2}x-1)^2 - (\frac{3}{4}x + 2)^2 = 0} \atop {x \geq -\frac{8}{3} }} \right. => \left \{ {{(\frac{1}{2}x-1 -(\frac{3}{4}x + 2))(\frac{1}{2}x-1 +\frac{3}{4}x + 2)=0 } \atop {x \geq -\frac{8}{3}}} \right. =>\\ \left \{ {{(-\frac{1}{4}x-3)(\frac{5}{4} x+1)= 0} \atop {x\geq-\frac{8}{3} }} \right. =>" align="absmiddle" class="latex-formula">
\left \{ {{(x+12)(x + \frac{4}{5})=0} \atop {x \geq-\frac{8}{3} }} \right." alt="=> \left \{ {{(x+12)(x + \frac{4}{5})=0} \atop {x \geq-\frac{8}{3} }} \right." align="absmiddle" class="latex-formula">
Теперь очевидно, что корни верхнего уравнения системы либо x = -12, либо x = 
. Первый корень не подходит по ОДЗ. Значит, ответ x = .
Ответ: x = .