-------------------ko chép đề nha---------
\(\Leftrightarrow\frac{4\left(x^2-4x+4\right)-3\left(2x+1\right)}{12}=\frac{12-2\left(4x^2-12x+9\right)}{12}\)
\(\Rightarrow4x^2+16x+16-6x-3=12-8x^2+24x-18\)
\(\Leftrightarrow4x^2+10x+13=-8x^2+24x-6\)
\(\Leftrightarrow4x^2+8x^2+10x-24x+13+6=0\)
\(\Leftrightarrow12x-14x+19=0\)
Ta có :\(\Delta'=7^2-12.19=-179< 0\)
\(\Rightarrow\)phương trình vô nghiệm