Ta có : \(2x\left(8x-1\right)^2\left(4x-1\right)=9\)
=> \(\left(8x^2-2x\right)\left(64x^2-16x+1\right)=9\)
=> \(512x^4-128x^3+8x^2-128x^3+32x^2-2x-9=0\)
=> \(512x^4-256x^3+40x^2-2x-9=0\)
=> \(512x^4+128x^3-384x^3-96x^2+136x^2+34x-36x-9=0\)
=> \(128x^3\left(4x+1\right)-96x^2\left(4x+1\right)+34x\left(4x+1\right)-9\left(4x+1\right)=0\)
=> \(\left(4x+1\right)\left(128x^3-96x^2+34x-9\right)=0\)
=> \(\left(4x+1\right)\left(128x^3-64x^2-32x^2+16x+18x-9\right)=0\)
=> \(\left(4x+1\right)\left(64x^2\left(2x-1\right)-16x\left(2x-1\right)+9\left(2x-1\right)\right)=0\)
=> \(\left(4x+1\right)\left(2x-1\right)\left(64x^2-16x+9\right)=0\)
Ta thấy : \(64x^2-16x+9\)
\(=\left(64x^2-2.8.x+1\right)+8\)
\(=\left(8x-1\right)^2+8>0\)
=> \(\left(4x+1\right)\left(2x-1\right)=0\)
=> \(\left[{}\begin{matrix}4x+1=0\\2x-1=0\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x=-\frac{1}{4}\\x=\frac{1}{2}\end{matrix}\right.\)
Vậy phương trình có tập nghiệm là \(S=\left\{-\frac{1}{4};\frac{1}{2}\right\}\)
