\(\dfrac{3x-1}{2}-\left(x-\dfrac{1}{4}\right)=\dfrac{4x-9}{8}\rightarrow\dfrac{3x-1}{2}-x+\dfrac{1}{4}=\dfrac{4x-9}{8}\rightarrow\dfrac{4\left(3x-1\right)-8x+2}{8}=\dfrac{4x-9}{8}\rightarrow4\left(3x-1\right)-8x+2=4x-9\rightarrow12x-4-8x+2=4x-9\rightarrow4x-2=4x-9\)
\(\dfrac{3x-1}{2}-\left(x-\dfrac{1}{4}\right)=\dfrac{4x-9}{8}\)
\(\Leftrightarrow\dfrac{3x-1}{2}-x+\dfrac{1}{4}=\dfrac{4x-9}{8}\)
\(\Leftrightarrow\dfrac{4\left(3x-1\right)-8x+2}{8}=\dfrac{4x-9}{8}\)
\(\Leftrightarrow4\left(3x-1\right)-8x+2=4x-9\)
\(\Leftrightarrow12x-4-8x+2=4x-9\)
\(\Leftrightarrow12x-8x-4x=-9+4-2\)
\(\Leftrightarrow0x=-7\)(vô lý)
Vậy S=\(\left(\varnothing\right)\)
\(\rightarrow4x-4x=-9+2\rightarrow0x=-7\)(Vô lý)
Vậy phương trình vô nghiệm
