\(\left(3x-1\right)\left(4x-1\right)\left(5x-1\right)-120=0\)
\(\left(3x-1\right)\left(4x-1\right)\left(5x-1\right)=120\)
\(\Leftrightarrow\left(18x^2-9x+1\right)\left(20x^2-90x+1\right)\)
Đặt \(a=19x^2-9x+1\) với \(a>0\) ta có \(PT:\left(a-1\right)\left(a+1\right)\)
\(\Leftrightarrow a^2-1=120\)
\(\Leftrightarrow a^2=121\)
\(\Leftrightarrow a=11\left(a>0\right)\)
Với \(a=11\) Ta có \(PT:19x^2-9x-10=0\)
\(\Leftrightarrow\left(10x+19\right)\left(x-1\right)=0\)
\(\Leftrightarrow10x+19=0\) Hoặc \(x-1=0\)
\(\Leftrightarrow10x=-19\) \(x=1\)
\(\Leftrightarrow x=-1.9\)
Vậy \(x=1\left(x\in Z\right)\)
