Ta có: (3x-3)(5-21x)+(7x+4)(6x-5)=45
\(\Leftrightarrow15x-63x^2-15+63x+42x^2-35x+24x-20-45=0\)
\(\Leftrightarrow-21x^2+67x-80=0\)
\(\Leftrightarrow-21\left(x^2-\frac{67}{21}x-\frac{80}{21}\right)=0\)
mà -21<0
nên \(x^2-\frac{67}{21}x-\frac{80}{21}=0\)
\(\Leftrightarrow x^2-2\cdot x\cdot\frac{67}{42}+\frac{4489}{1764}-\frac{11209}{1764}=0\)
\(\Leftrightarrow\left(x-\frac{67}{42}\right)^2=\frac{11209}{1764}\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\frac{67}{42}=\frac{\sqrt{11209}}{42}\\x-\frac{67}{42}=-\frac{\sqrt{11209}}{42}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{67+\sqrt{11209}}{42}\\x=\frac{67-\sqrt{11209}}{42}\end{matrix}\right.\)
Vậy: \(x\in\left\{\frac{67+\sqrt{11209}}{42};\frac{67-\sqrt{11209}}{42}\right\}\)