Đáp án:

\(\begin{array}{l}
1,\\
a,\,\,\,24x - 16\\
b,\,\,\, - 3{x^2} + 15x - 19\\
c,\,\,\,6{x^2}y + 2{y^3}\\
d,\,\,\,5 - 10x{y^2}\\
2,\\
a,\,\,\,A = 7x\\
b,\,\,\,B = 16{x^2}\\
3,\\
a,\,\,\,x = \dfrac{{17}}{{59}}\\
b,\,\,\,\left[ \begin{array}{l}
x = 0\\
x =  - 4
\end{array} \right.\\
c,\,\,x = 1\\
d,\,\,x = 1
\end{array}\)

Giải thích các bước giải:

 Ta có:

\(\begin{array}{l}
1,\\
a,\\
9{x^2} - {\left( {3x - 4} \right)^2}\\
 = {\left( {3x} \right)^2} - {\left( {3x - 4} \right)^2}\\
 = \left[ {3x - \left( {3x - 4} \right)} \right].\left[ {3x + \left( {3x - 4} \right)} \right]\\
 = \left( {3x - 3x + 4} \right)\left( {3x + 3x - 4} \right)\\
 = 4.\left( {6x - 4} \right)\\
 = 24x - 16\\
b,\\
{\left( {x - 3} \right)^3} + {\left( {2 - x} \right)^3}\\
 = \left( {{x^3} - 3.{x^2}.3 + 3.x{{.3}^2} - {3^3}} \right) + \left( {{2^3} - {{3.2}^2}.x + 3.2.{x^2} - {x^3}} \right)\\
 = \left( {{x^3} - 9{x^2} + 27x - 27} \right) + \left( {8 - 12x + 6{x^2} - {x^3}} \right)\\
 = \left( {{x^3} - {x^3}} \right) + \left( { - 9{x^2} + 6{x^2}} \right) + \left( {27x - 12x} \right) + \left( { - 27 + 8} \right)\\
 =  - 3{x^2} + 15x - 19\\
c,\\
{\left( {x + y} \right)^3} - {\left( {x - y} \right)^3}\\
 = \left( {{x^3} + 3{x^2}y + 3x{y^2} + {y^3}} \right) - \left( {{x^3} - 3{x^2}y + 3x{y^2} - {y^3}} \right)\\
 = {x^3} + 3{x^2}y + 3x{y^2} + {y^3} - {x^3} + 3{x^2}y - 3x{y^2} + {y^3}\\
 = 6{x^2}y + 2{y^3}\\
d,\\
{\left( {3 - x{y^2}} \right)^2} - {\left( {2 + x{y^2}} \right)^3}\\
 = \left[ {\left( {3 - x{y^2}} \right) - \left( {2 + x{y^2}} \right)} \right].\left[ {\left( {3 - x{y^2}} \right) + \left( {2 + x{y^2}} \right)} \right]\\
 = \left( {3 - x{y^2} - 2 - x{y^2}} \right).\left( {3 - x{y^2} + 2 + x{y^2}} \right)\\
 = \left( {1 - 2x{y^2}} \right).5\\
 = 5 - 10x{y^2}\\
2,\\
a,\\
A = {\left( {x + 1} \right)^2} - {\left( {x - 1} \right)^2} + 3{x^3} - 3x\left( {x + 1} \right)\left( {x - 1} \right)\\
 = \left( {{x^2} + 2x.1 + {1^2}} \right) - \left( {{x^2} - 2.x.1 + {1^2}} \right) + 3{x^3} - 3x.\left( {{x^2} - {1^2}} \right)\\
 = \left( {{x^2} + 2x + 1} \right) - \left( {{x^2} - 2x + 1} \right) + 3{x^3} - 3x.\left( {{x^2} - 1} \right)\\
 = {x^2} + 2x + 1 - {x^2} + 2x - 1 + 3{x^3} - 3{x^3} + 3x\\
 = \left( {3{x^3} - 3{x^3}} \right) + \left( {{x^2} - {x^2}} \right) + \left( {2x + 2x + 3x} \right) + \left( {1 - 1} \right)\\
 = 7x\\
b,\\
B = {\left( {2x + 1} \right)^2} + 2.\left( {4{x^2} - 1} \right) + {\left( {2x - 1} \right)^2}\\
 = {\left( {2x + 1} \right)^2} + 2.\left[ {{{\left( {2x} \right)}^2} - {1^2}} \right] + {\left( {2x - 1} \right)^2}\\
 = {\left( {2x + 1} \right)^2} + 2.\left( {2x + 1} \right)\left( {2x - 1} \right) + {\left( {2x - 1} \right)^2}\\
 = {\left[ {\left( {2x + 1} \right) + \left( {2x - 1} \right)} \right]^2}\\
 = {\left( {2x + 1 + 2x - 1} \right)^2}\\
 = {\left( {4x} \right)^2}\\
 = 16{x^2}\\
3,\\
a,\\
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} + 9x + 2x + 6} \right) = 38\\
 \Leftrightarrow 3.\left( { - 4{x^2} + 5x - 1} \right) + 4.\left( {3{x^2} + 11x + 6} \right) = 38\\
 \Leftrightarrow  - 12{x^2} + 15x - 3 + 12{x^2} + 44x + 24 = 38\\
 \Leftrightarrow \left( { - 12{x^2} + 12{x^2}} \right) + \left( {15x + 44x} \right) + \left( { - 3 + 24} \right) = 38\\
 \Leftrightarrow 59x + 21 = 38\\
 \Leftrightarrow 59x = 17\\
 \Leftrightarrow x = \dfrac{{17}}{{59}}\\
b,\\
\left( {x - 2} \right)\left( {x - 1} \right) = x.\left( {2x + 1} \right) + 2\\
 \Leftrightarrow {x^2} - x - 2x + 2 = 2{x^2} + x + 2\\
 \Leftrightarrow {x^2} - 3x + 2 - 2{x^2} - x - 2 = 0\\
 \Leftrightarrow  - {x^2} - 4x = 0\\
 \Leftrightarrow  - x.\left( {x + 4} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
x = 0\\
x + 4 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 0\\
x =  - 4
\end{array} \right.\\
c,\\
\left( {2x - 1} \right)\left( {{x^2} - x + 1} \right) = 2{x^3} - 3{x^2} + 2\\
 \Leftrightarrow 2x.\left( {{x^2} - x + 1} \right) - 1.\left( {{x^2} - x + 1} \right) = 2{x^3} - 3{x^2} + 2\\
 \Leftrightarrow 2{x^3} - 2{x^2} + 2x - {x^2} + x - 1 - 2{x^3} + 3{x^2} - 2 = 0\\
 \Leftrightarrow \left( {2{x^3} - 2{x^3}} \right) + \left( { - 2{x^2} - {x^2} + 3{x^2}} \right) + \left( {2x + x} \right) + \left( { - 1 - 2} \right) = 0\\
 \Leftrightarrow 3x - 3 = 0\\
 \Leftrightarrow 3x = 3\\
 \Leftrightarrow x = 1\\
d,\\
\left( {x + 1} \right)\left( {x + 2} \right)\left( {x + 5} \right) - {x^3} - 8{x^2} = 27\\
 \Leftrightarrow \left( {{x^2} + 2x + x + 2} \right)\left( {x + 5} \right) - {x^3} - 8{x^2} = 27\\
 \Leftrightarrow \left( {{x^2} + 3x + 2} \right)\left( {x + 5} \right) - {x^3} - 8{x^2} = 27\\
 \Leftrightarrow \left( {{x^2} + 3x + 2} \right).x + \left( {{x^2} + 3x + 2} \right).5 - {x^3} - 8{x^2} = 27\\
 \Leftrightarrow {x^3} + 3{x^2} + 2x + 5{x^2} + 15x + 10 - {x^3} - 8{x^2} - 27 = 0\\
 \Leftrightarrow \left( {{x^3} - {x^3}} \right) + \left( {3{x^2} + 5{x^2} - 8{x^2}} \right) + \left( {2x + 15x} \right) + \left( {10 - 27} \right) = 0\\
 \Leftrightarrow 17x - 17 = 0\\
 \Leftrightarrow 17x = 17\\
 \Leftrightarrow x = 1
\end{array}\)