\(c.\dfrac{1}{4}x+\dfrac{4}{7}=\dfrac{1}{6}x-\dfrac{3}{7}\\ \dfrac{1}{4}x-\dfrac{1}{6}x=-\dfrac{3}{7}-\dfrac{4}{7}\\ \dfrac{1}{12}x=-1\\ x=-1:\dfrac{1}{12}\\ x=-12\\ d.\left(3x+\dfrac{1}{5}\right)^2-\dfrac{16}{25}=0\\ \left(3x+\dfrac{1}{5}\right)^2=\dfrac{16}{25}\\ \left(3x+\dfrac{1}{5}\right)^2=\left(\dfrac{4}{5}\right)^2\\ TH1:3x+\dfrac{1}{5}=\dfrac{4}{5}\\ 3x=\dfrac{4}{5}-\dfrac{1}{5}=\dfrac{3}{5}\\ x=\dfrac{3}{5}:3=\dfrac{1}{5}\\ TH2:3x+\dfrac{1}{5}=-\dfrac{4}{5}\\ 3x=-\dfrac{4}{5}-\dfrac{1}{5}=-1\\ x=-\dfrac{1}{3}\\ e.\dfrac{x+1}{2}+\dfrac{x+1}{3}=\dfrac{x+1}{4}+\dfrac{x+1}{5}\\ \dfrac{x+1}{2}+\dfrac{x+1}{3}-\dfrac{x+1}{4}-\dfrac{x+1}{5}=0\\ \left(x+1\right)\left(\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}-\dfrac{1}{5}\right)\\ x+1=0\\ x=-1\)
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