a) Ta có: \(3x\left(x+1\right)-2x\left(x+20\right)=-1-x\)
\(\Leftrightarrow3x^2+3x-2x^2-40x+1+x=0\)
\(\Leftrightarrow x^2-36x+1=0\)
\(\Leftrightarrow x^2-36x+324-323=0\)
\(\Leftrightarrow\left(x-18\right)^2=323\)
\(\Leftrightarrow\left[{}\begin{matrix}x-18=\sqrt{323}\\x-18=-\sqrt{323}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=18+\sqrt{323}\\x=18-\sqrt{323}\end{matrix}\right.\)
Vậy: \(x\in\left\{18+\sqrt{323};18-\sqrt{323}\right\}\)
b) Ta có: \(\left(3x-1\right)\left(2x+7\right)-\left(x+1\right)\left(6x-5\right)=16\)
\(\Leftrightarrow6x^2+21x-2x-7-\left(6x^2-5x+6x-5\right)-16=0\)
\(\Leftrightarrow6x^2+19x-7-\left(6x^2+x-5\right)-16=0\)
\(\Leftrightarrow6x^2+19x-7-6x^2-x+5-16=0\)
\(\Leftrightarrow18x-18=0\)
\(\Leftrightarrow18x=18\)
hay x=1
Vậy: x=1
c) Ta có: \(\left(10x+9\right)\cdot x-\left(5x-1\right)\left(2x+3\right)=8\)
\(\Leftrightarrow10x^2+9x-\left(10x^2+15x-2x-3\right)-8=0\)
\(\Leftrightarrow10x^2+9x-10x^2-13x+3-8=0\)
\(\Leftrightarrow-4x-5=0\)
\(\Leftrightarrow-4x=5\)
hay \(x=\frac{-5}{4}\)
Vậy: \(x=\frac{-5}{4}\)