\(a.\left(1\dfrac{3}{4}-x\right)\cdot2\dfrac{2}{5}=4,5\\ \left(\dfrac{7}{4}-x\right)\cdot\dfrac{12}{5}=\dfrac{9}{2}\\ \dfrac{7}{4}-x=\dfrac{9}{2}:\dfrac{12}{5}=\dfrac{45}{24}\\ x=\dfrac{7}{4}-\dfrac{45}{24}=-\dfrac{3}{24}=-\dfrac{1}{8}\\ b.60\%x+\dfrac{2}{3}x=\dfrac{1}{3}\cdot6\dfrac{1}{3}\\ \dfrac{3}{5}\cdot x+\dfrac{2}{3}\cdot x=\dfrac{1}{3}\cdot\dfrac{19}{3}\\ x\cdot\left(\dfrac{3}{5}+\dfrac{2}{3}\right)=\dfrac{19}{9}\\ x\cdot\dfrac{19}{15}=\dfrac{19}{9}\\ x=\dfrac{19}{9}:\dfrac{19}{5}=\dfrac{5}{9}\\ c.\left(5x+3\right)^{28}=16^7\\ \left(5x+3\right)^{28}=\left(2^4\right)^7\\ \left(5x+3\right)^{28}=2^{28}\\ TH1:5x+3=2\\ 5x=2-3=-1\\ x=-\dfrac{1}{5}\\ TH2:5x+3=-2\\ 5x=-2-3=-5\\ x=\dfrac{-5}{5}=-1\\ d.\left(\dfrac{2}{3}\right)^{5x-1}=\left(-\dfrac{4}{9}\right)^8\\ \left(\dfrac{2}{3}\right)^{5x-1}=\left(\dfrac{4}{9}\right)^8\\ \left(\dfrac{2}{3}\right)^{5x-1}=\left[\left(\dfrac{2}{3}\right)^2\right]^8\\ \left(\dfrac{2}{3}\right)^{5x-1}=\left(\dfrac{2}{3}\right)^{16}\\ 5x-1=16\\ 5x=16+1=17\\ x=\dfrac{17}{5}\)