Ta có: \(2x\left(8x-1\right)^2\cdot\left(4x-1\right)=9\)

\(\Leftrightarrow\left(8x-1\right)^2\cdot\left(8x^2-2x\right)=9\)

\(\Leftrightarrow\left(64x^2-16x+1\right)\left(8x^2-2x\right)-9=0\)

\(\Leftrightarrow512x^4-128x^3-128x^3+32x^2+8x^2-2x-9=0\)

\(\Leftrightarrow512x^4-256x^3+40x^2-2x-9=0\)

\(\Leftrightarrow256x^3\left(2x-1\right)+40x^2-20x+18x-9=0\)

\(\Leftrightarrow256x^3\left(2x-1\right)+20x\left(2x-1\right)+9\left(2x-1\right)=0\)

\(\Leftrightarrow\left(2x-1\right)\left(256x^3+20x+9\right)=0\)

\(\Leftrightarrow\left(2x-1\right)\left(256x^3+64x^2-64x^2-16x+36x+9\right)=0\)

\(\Leftrightarrow\left(2x-1\right)\left[64x^2\left(4x+1\right)-4x\left(4x+1\right)+9\left(4x+1\right)\right]=0\)

\(\Leftrightarrow\left(2x-1\right)\left(4x+1\right)\left(64x^2-4x+9\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-1=0\\4x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=1\\4x=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=-\dfrac{1}{4}\end{matrix}\right.\)

Vậy: \(x\in\left\{\dfrac{1}{2};-\dfrac{1}{4}\right\}\) 

nhân phá tung ra là xog mà

Giups mik vs

=>x=1

\(a,x\left(8x-2\right)-8x^2+12=0\)

\(\Rightarrow8x^2-2x-8x^2+12=0\)

\(\Rightarrow-2x+12=0\)

\(\Rightarrow-2x=-12\)

\(\Rightarrow x=6\)

\(b,x\left(4x-5\right)-\left(2x+1\right)^2=0\)

\(\Rightarrow4x^2-5x-4x^2-4x-1=0\)

\(\Rightarrow-9x-1=0\)

\(\Rightarrow-9x=1\)

\(\Rightarrow x=\frac{-1}{9}\)

a) x(8 - 2) - 8x² + 12 = 0

x(8 - 2) - 8x² = 12 - 0

x(8 - 2) - 8x² = 12

2x = 12

x = 6

b) x(4x - 5) - (2x + 1)² = 0

9x - 1 = 0

9x = 0 + 1

9x = 1

x = -1/9

\(a,x\left(8x-2\right)-8x^2+12=0\)

\(\Leftrightarrow8x^2-2x-8x^2+12=0\)

\(\Leftrightarrow-2x+12=0\)

\(\Leftrightarrow-2x=-12\)

\(\Leftrightarrow x=6\)

\(b,x\left(4x-5\right)-\left(2x+1\right)^2=0\)

\(\Leftrightarrow4x^2-5x-4x^2-4x-1=0\)

\(\Leftrightarrow-9x-1=0\)

\(\Leftrightarrow-9x=1\)

\(\Leftrightarrow x=-\frac{1}{9}\)