b) 

\(\left(8x-7\right)\left(8x-5\right)\left(2x-1\right)\left(4x-1\right)=9\)

\(\Leftrightarrow\left(8x-7\right)\left(8x-5\right)\left(4x-4\right)\left(8x-2\right)=72\)

Đặt a = 8x - 5, ta được:

\(\left(a-2\right).a\left(a+1\right)\left(a+3\right)=72\)

\(\Leftrightarrow a^4+4a^3+3a^2-2a^3-8a^2-6a-72=0\)

\(\Leftrightarrow a^4+4a^3-2a^3-8a^2+3a^2+12a-18a-72=0\)\(\Leftrightarrow\left(a^4+4a^3\right)-\left(2a^3+8a^2\right)+\left(3a^2+12a\right)-\left(18a+72\right)=0\)

\(\Leftrightarrow a^3\left(a+4\right)-2a^2\left(a+4\right)+3a\left(a+4\right)-18\left(a+4\right)=0\) \(\Leftrightarrow\left(a+4\right)\left(a^3-2a^2+3a-18\right)=0\)

\(\Leftrightarrow\left(a+4\right)\left(a^3-3a^2+a^2-3a+6a-18\right)=0\)

\(\Leftrightarrow\left(a+4\right)\left[\left(a^3-3a^2\right)+\left(a^2-3a\right)+\left(6a-18\right)\right]=0\)

\(\Leftrightarrow\left(a+4\right)\left[a^2\left(a-3\right)+a\left(a-3\right)+6\left(a-3\right)\right]=0\)

\(\Leftrightarrow\left(a+4\right)\left(a-3\right)\left(a^2+a+6\right)=0\)

Ta có: \(a^2+a+6=a^2+2.a.\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{23}{4}=\left(a+\dfrac{1}{2}\right)^2+\dfrac{23}{4}\)

\(\left(a+\dfrac{1}{2}\right)\ge0\)

Suy ra \(\left(a+\dfrac{1}{2}\right)^2+\dfrac{23}{4}>0\)

=> a2 +a+6 = 0 (loại)

Suy ra: a = -4 hoặc a=3

Với a = -4, ta được:

8x - 5 = -4

=> x = \(\dfrac{1}{8}\)

Với a = 3, ta được:

8x - 5 = 3

=> x = 1