a)
\(\dfrac{2}{5x-2}+\dfrac{4}{1-5x}=\dfrac{2}{\left(5x-2\right)\left(5x-1\right)}\left(x\ne\dfrac{2}{5};x\ne\dfrac{1}{5}\right)\\ \Leftrightarrow\dfrac{2}{5x-2}-\dfrac{4}{5x-1}=\dfrac{2}{\left(5x-2\right)\left(5x-1\right)}\\ \Leftrightarrow\dfrac{2\left(5x-1\right)}{\left(5x-2\right)\left(5x-1\right)}-\dfrac{4\left(5x-2\right)}{\left(5x-2\right)\left(5x-1\right)}=\dfrac{2}{\left(5x-2\right)\left(5x-1\right)}\\ \Leftrightarrow2\left(5x-1\right)-4\left(5x-2\right)=2\\ \Leftrightarrow10x-2-20x+8=2\\ \Leftrightarrow-10x+6=2\\ \Leftrightarrow10x=6-2\\ \Leftrightarrow10x=4\\ \Leftrightarrow x=\dfrac{2}{5}\left(ktm\right)\)
b)
\(\dfrac{2}{2x+1}+\dfrac{1}{x+1}=\dfrac{3}{\left(2x+1\right)\left(x+1\right)}\left(x\ne-\dfrac{1}{2};x\ne-1\right)\\ \Leftrightarrow\dfrac{2\left(x+1\right)}{\left(2x+1\right)\left(x+1\right)}+\dfrac{2x+1}{\left(2x+1\right)\left(x+1\right)}=\dfrac{3}{\left(2x+1\right)\left(x+1\right)}\\ \Leftrightarrow2\left(x+1\right)+2x+1=3\\ \Leftrightarrow2x+2+2x+1=3\\ \Leftrightarrow4x+3=3\\ \Leftrightarrow4x=0\\ \Leftrightarrow x=0\left(tm\right)\)
c)
\(\dfrac{3}{x-2}+\dfrac{2}{x+1}=\dfrac{2x-7}{x^2-x-2}\left(x\ne2;x\ne-1\right)\\ \Leftrightarrow\dfrac{3}{x-2}+\dfrac{2}{x+1}=\dfrac{2x-7}{\left(x-2\right)\left(x+1\right)}\\ \Leftrightarrow\dfrac{3\left(x+1\right)}{\left(x-2\right)\left(x+1\right)}+\dfrac{2\left(x-2\right)}{\left(x-2\right)\left(x+1\right)}=\dfrac{2x-7}{\left(x-2\right)\left(x+1\right)}\\ \Leftrightarrow3\left(x+1\right)+2\left(x-2\right)=2x-7\\ \Leftrightarrow3x+3+2x-4=2x-7\\ \Leftrightarrow5x-1=2x-7\\ \Leftrightarrow5x-2x=-7+1\\ \Leftrightarrow3x=-6\\ \Leftrightarrow x=-2\left(tm\right)\)