\(\dfrac{8x^2}{3\left(1-4x^2\right)}=\dfrac{2x}{6x-3}-\dfrac{1+8x}{4+8x}\left(ĐK:x\ne\pm\dfrac{1}{2}\right)\\ \Leftrightarrow\dfrac{8x^2}{-3\left[\left(2x\right)^2-1^2\right]}=\dfrac{2x}{3\left(2x-1\right)}-\dfrac{8x+1}{4\left(1+2x\right)}\\ \Leftrightarrow-\dfrac{8x^2}{3\left(2x-1\right)\left(2x+1\right)}=\dfrac{2x}{3\left(2x-1\right)}-\dfrac{8x+1}{4\left(2x+1\right)}\\ \Leftrightarrow-\dfrac{32x^2}{12\left(2x-1\right)\left(2x+1\right)}=\dfrac{8x\left(2x+1\right)}{12\left(2x-1\right)\left(2x+1\right)}-\dfrac{3\left(8x+1\right)\left(2x-1\right)}{12\left(2x+1\right)\left(2x-1\right)}\\ \)
\(\Leftrightarrow\dfrac{-32x^2}{12\left(2x-1\right)\left(2x+1\right)}=\dfrac{16x^2+8x-3\left(16x^2+2x-8x-1\right)}{12\left(2x+1\right)\left(2x-1\right)}\\ \Rightarrow-32x^2=16x^2+8x-3\left(16x^2-6x-1\right)\\ \Leftrightarrow-32x^2=16x^2+8x-48x^2+18x+3\\ \Leftrightarrow-32x^2=-32x^2+26x+3\\ \Leftrightarrow32x^2-32x^2+26x+3=0\\ \Leftrightarrow26x+3=0\\ \Leftrightarrow26x=-3\\ \Leftrightarrow x=-\dfrac{3}{26}\left(TMDK\right)\)
Vậy pt có nghiệm duy nhất: \(x=-\dfrac{3}{26}\)
\(\dfrac{8}{3\left(1-4x^2\right)}=\dfrac{2x}{6x-3}-\dfrac{1+8x}{4+8x}\)
=>\(\dfrac{-8}{3\left(2x-1\right)\left(2x+1\right)}=\dfrac{2x}{3\left(2x-1\right)}-\dfrac{8x+1}{4\left(2x+1\right)}\)
=>\(\dfrac{-32}{12\left(2x-1\right)\left(2x+1\right)}=\dfrac{2x\cdot4\left(2x+1\right)-3\cdot\left(2x-1\right)\left(8x+1\right)}{12\left(2x-1\right)\left(2x+1\right)}\)
=>\(8x\left(2x+1\right)-3\left(16x^2-6x-1\right)=-32\)
=>\(16x^2+8x-48x^2+18x+3+32=0\)
=>\(-32x^2+26x+35=0\)
\(\text{Δ}=26^2-4\cdot\left(-32\right)\cdot35=5156>0\)
=>Phương trình có hai nghiệm phân biệt là:
\(\left[{}\begin{matrix}x=\dfrac{-26-\sqrt{5156}}{2\cdot\left(-32\right)}=\dfrac{26+2\sqrt{1289}}{64}=\dfrac{13-\sqrt{1289}}{32}\left(nhận\right)\\x=\dfrac{13+\sqrt{1289}}{32}\left(nhận\right)\end{matrix}\right.\)