Đáp án:

$\begin{array}{l}
1)\dfrac{1}{4}{x^2} - \left( {\dfrac{1}{2}x - 4} \right).\dfrac{1}{2}x =  - 14\\
 \Leftrightarrow \dfrac{1}{4}{x^2} - \dfrac{1}{4}{x^2} - 2x =  - 14\\
 \Leftrightarrow  - 2x =  - 14\\
 \Leftrightarrow x = 7\\
\text{Vậy}\,x = 7\\
2)3\left( {1 - 4x} \right) + 4\left( {3x - 2} \right)\left( {x + 3} \right) =  - 27\\
 \Leftrightarrow 3 - 12x + 4.\left( {3{x^2} + 9x - 2x - 6} \right) =  - 27\\
 \Leftrightarrow 3 - 12x + 12{x^2} + 28x - 24 + 27 = 0\\
 \Leftrightarrow 12{x^2} + 16x + 6 = 0\\
 \Leftrightarrow 3{x^2} + 4x + 4 = 0\left( {vn} \right)\\
\text{Vậy pt vô nghiệm}\\
3)\left( {x + 3} \right)\left( {{x^2} - 3x + 9} \right) - x\left( {x - 1} \right)\left( {x + 1} \right) = 27\\
 \Leftrightarrow {x^3} + 27 - x\left( {{x^2} - 1} \right) = 27\\
 \Leftrightarrow {x^3} - {x^3} + x = 0\\
 \Leftrightarrow x = 0\\
\text{Vậy}\,x = 0\\
4)6x\left( {5x + 3} \right) + 3x\left( {1 - 10x} \right) = 7\\
 \Leftrightarrow 30{x^2} + 18x + 3x - 30{x^2} = 7\\
 \Leftrightarrow 21x = 7\\
 \Leftrightarrow x = \dfrac{1}{3}\\
\text{Vậy}\,x = \dfrac{1}{3}\\
5)\left( {3x - 3} \right)\left( {5 - 21x} \right) + \left( {7x + 4} \right)\left( {9x - 5} \right) = 44\\
 \Leftrightarrow 15x - 63{x^2} - 15 + 63x + 63{x^2} - 35x + 36x - 20 = 44\\
 \Leftrightarrow 79x = 44 + 15 + 20\\
 \Leftrightarrow x = 1\\
\text{Vậy}\,x = 1\\
6)3\left( {1 - 4x} \right)\left( {x - 1} \right) + 4\left( {3x + 2} \right)\left( {x + 3} \right) = 38\\
 \Leftrightarrow 3.\left( {x - 1 - 4{x^2} + 4x} \right) + 4.\left( {3{x^2} + 11x + 6} \right) = 38\\
 \Leftrightarrow  - 12{x^2} + 15x - 3 + 12{x^2} + 44x + 24 = 38\\
 \Leftrightarrow 59x = 17\\
 \Leftrightarrow x = \dfrac{{17}}{{59}}\\
\text{Vậy}\,x = \dfrac{{17}}{{59}}\\
7)5.\left( {2x + 3} \right)\left( {x + 2} \right) - 2\left( {5x - 4} \right)\left( {x - 1} \right) = 75\\
 \Leftrightarrow 5.\left( {2{x^2} + 7x + 6} \right) - 2.\left( {5{x^2} - 9x + 4} \right) = 75\\
 \Leftrightarrow 10{x^2} + 35x + 30 - 10{x^2} + 18x - 8 = 75\\
 \Leftrightarrow 53x = 53\\
 \Leftrightarrow x = 1\\
\text{Vậy}\,x = 1\\
8)2{x^2} + 3\left( {x - 1} \right)\left( {x + 1} \right) = 5x\left( {x + 1} \right)\\
 \Leftrightarrow 2{x^2} + 3\left( {{x^2} - 1} \right) = 5{x^2} + 5x\\
 \Leftrightarrow  - 3 = 5x\\
 \Leftrightarrow x = \dfrac{{ - 3}}{5}\\
\text{Vậy}\,x = \dfrac{{ - 3}}{5}\\
9)\left( {8 - 5x} \right)\left( {x + 2} \right) + 4\left( {x + 2} \right)\left( {x + 1} \right)\\
 + 2\left( {x - 2} \right)\left( {x + 2} \right) = 0\\
 \Leftrightarrow 8x + 16 - 5{x^2} - 10x + 4\left( {{x^2} + 3x + 2} \right)\\
 + 2\left( {{x^2} - 4} \right) = 0\\
 \Leftrightarrow  - 5{x^2} - 2x + 16 + 4{x^2} + 12x + 8 + 2{x^2} - 8 = 0\\
 \Leftrightarrow {x^2} + 10x + 16 = 0\\
 \Leftrightarrow \left( {x + 2} \right)\left( {x + 8} \right) = 0\\
 \Leftrightarrow x =  - 2;x =  - 8\\
\text{Vậy}\,x =  - 2;x =  - 8\\
10)\left( {x - 2} \right)\left( {x - 1} \right) = x\left( {2x + 1} \right) + 2\\
 \Leftrightarrow {x^2} - 3x + 2 = 2{x^2} + x + 2\\
 \Leftrightarrow {x^2} + 4x = 0\\
 \Leftrightarrow x\left( {x + 4} \right) = 0\\
 \Leftrightarrow x = 0;x =  - 4\\
\text{Vậy}\,x = 0;x =  - 4
\end{array}$