Giải thích các bước giải:

Ta có:

\(\begin{array}{l}
d,\\
2x.\left( {1 - 3x} \right) + 6{x^2} - 7x - 10 = 0\\
 \Leftrightarrow \left( {2x - 6{x^2}} \right) + 6{x^2} - 7x - 10 = 0\\
 \Leftrightarrow 2x - 6{x^2} + 6{x^2} - 7x - 10 = 0\\
 \Leftrightarrow  - 5x - 10 = 0\\
 \Leftrightarrow 5x =  - 10\\
 \Leftrightarrow x =  - 2\\
e,\\
\left( {x + 2} \right)\left( {3x - 1} \right) - 3.{\left( {x - 3} \right)^2} = 7x + 5\\
 \Leftrightarrow \left( {3{x^2} - x + 6x - 2} \right) - 3.\left( {{x^2} - 6x + 9} \right) - 7x - 5 = 0\\
 \Leftrightarrow 3{x^2} + 5x - 2 - 3{x^2} + 18x - 27 - 7x - 5 = 0\\
 \Leftrightarrow 16x - 34 = 0\\
 \Leftrightarrow 16x = 34\\
 \Leftrightarrow x = \frac{{17}}{8}\\
f,\\
\left( {2x + 1} \right)\left( {x - 2} \right) + 4x = {x^2} + 4\\
 \Leftrightarrow \left( {2x + 1} \right)\left( {x - 2} \right) + 4x - {x^2} - 4 = 0\\
 \Leftrightarrow \left( {2x + 1} \right)\left( {x - 2} \right) - \left( {{x^2} - 4x + 4} \right) = 0\\
 \Leftrightarrow \left( {2x + 1} \right)\left( {x - 2} \right) - {\left( {x - 2} \right)^2} = 0\\
 \Leftrightarrow \left( {x - 2} \right).\left[ {\left( {2x + 1} \right) - \left( {x - 2} \right)} \right] = 0\\
 \Leftrightarrow \left( {x - 2} \right)\left( {x + 3} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
x - 2 = 0\\
x + 3 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 2\\
x =  - 3
\end{array} \right.
\end{array}\) 

`d) => 2x-6x^2+6x^2-7x-10=0`

`=> -5x-10=0`

`=> 5x=-10`

`=> x=-2`

`e) => 3x^2-x+6x-2 -3(x^2-6x+9)-7x-5=0`

`=> 3x^2+5x-2-3x^2+18x-27-7x-5=0`

`=> 16x-34=0`

`=> 16x=34`

`=> x=34/16=17/8`

`f) => (2x+1)(x-2)+4x^2-x^2-4=0`

`=> (2x+1)(x-2)-(x-2)^2=0`

`=> (x-2)(x+3)=0`

`=> x-2=0` hoặc `x+3=0`

`=> x=2` hoặc `x=-3`

`Huyền`