Đáp án:
\(\begin{array}{l}
Bài\,\,2:\\
1,\\
\left( {x - 3} \right).\left( {{x^2} - x + 9} \right)\\
2,\\
\left( {{x^2} + x + 1} \right)\left( {x + 1} \right)\\
3,\\
{\left( {x - 1} \right)^3}.\left( {x + 1} \right)\\
4,\\
{\left( {x + 1} \right)^2}.\left( {{x^2} + 1} \right)\\
5,\\
\left( {x - 2y - 2} \right).\left( {x + 2y} \right)\\
6,\\
\left( {{x^2} + 2x + 2} \right).\left( {{x^2} - 2} \right)\\
Bài\,\,\,3:\\
1,\\
\left( {x + 1} \right).\left( {x + 3} \right)\\
2,\\
\left( {x + 1} \right).\left( {x + 4} \right)\\
3,\\
\left( {x - 1} \right)\left( {x - 3} \right)\\
4,\\
\left( {x - 1} \right)\left( {x - 2} \right)\\
5,\\
\left( {x - 1} \right)\left( {x + 6} \right)\\
6,\\
\left( {x - 3} \right)\left( {x + 1} \right)\\
7,\\
\left( {x - 2} \right)\left( {x + 3} \right)\\
8,\\
\left( {x - 3} \right)\left( {x + 2} \right)\\
9,\\
\left( {3x + 2} \right).\left( {2x + 1} \right)\\
10,\\
\left( {x - 3} \right)\left( {3x - 2} \right)
\end{array}\)
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
Bài\,\,2:\\
1,\\
{x^3} - 4{x^2} + 12x - 27\\
= \left( {{x^3} - 3{x^2}} \right) + \left( { - {x^2} + 3x} \right) + \left( {9x + 37} \right)\\
= {x^2}.\left( {x - 3} \right) - x.\left( {x - 3} \right) + 9.\left( {x - 3} \right)\\
= \left( {x - 3} \right).\left( {{x^2} - x + 9} \right)\\
2,\\
{x^3} + 2{x^2} + 2x + 1\\
= \left( {{x^3} + {x^2} + x} \right) + \left( {{x^2} + x + 1} \right)\\
= x\left( {{x^2} + x + 1} \right) + \left( {{x^2} + x + 1} \right)\\
= \left( {{x^2} + x + 1} \right)\left( {x + 1} \right)\\
3,\\
{x^4} - 2{x^3} + 2x - 1\\
= \left( {{x^4} - 2{x^3} + {x^2}} \right) + \left( { - {x^2} + 2x - 1} \right)\\
= {x^2}.\left( {{x^2} - 2x + 1} \right) - \left( {{x^2} - 2x + 1} \right)\\
= \left( {{x^2} - 2x + 1} \right).\left( {{x^2} - 1} \right)\\
= \left( {{x^2} - 2.x.1 + {1^2}} \right).\left( {{x^2} - {1^2}} \right)\\
= {\left( {x - 1} \right)^2}.\left( {x - 1} \right).\left( {x + 1} \right)\\
= {\left( {x - 1} \right)^3}.\left( {x + 1} \right)\\
4,\\
{x^4} + 2{x^3} + 2{x^2} + 2x + 1\\
= \left( {{x^4} + 2{x^3} + {x^2}} \right) + \left( {{x^2} + 2x + 1} \right)\\
= {x^2}.\left( {{x^2} + 2x + 1} \right) + \left( {{x^2} + 2x + 1} \right)\\
= \left( {{x^2} + 2x + 1} \right).\left( {{x^2} + 1} \right)\\
= {\left( {x + 1} \right)^2}.\left( {{x^2} + 1} \right)\\
5,\\
{x^2} - 2x - 4{y^2} - 4y\\
= \left( {{x^2} - 2x + 1} \right) + \left( { - 4{y^2} - 4y - 1} \right)\\
= \left( {{x^2} - 2.x.1 + {1^2}} \right) - \left( {4{y^2} + 4y + 1} \right)\\
= {\left( {x - 1} \right)^2} - \left[ {{{\left( {2y} \right)}^2} + 2.2y.1 + {1^2}} \right]\\
= {\left( {x - 1} \right)^2} - {\left( {2y + 1} \right)^2}\\
= \left[ {\left( {x - 1} \right) - \left( {2y + 1} \right)} \right].\left[ {\left( {x - 1} \right) + \left( {2y + 1} \right)} \right]\\
= \left( {x - 1 - 2y - 1} \right).\left( {x - 1 + 2y + 1} \right)\\
= \left( {x - 2y - 2} \right).\left( {x + 2y} \right)\\
6,\\
{x^4} + 2{x^3} - 4x - 4\\
= \left( {{x^4} + 2{x^3} + 2{x^2}} \right) + \left( { - 2{x^2} - 4x - 4} \right)\\
= {x^2}.\left( {{x^2} + 2x + 2} \right) - 2.\left( {{x^2} + 2x + 2} \right)\\
= \left( {{x^2} + 2x + 2} \right).\left( {{x^2} - 2} \right)\\
Bài\,\,\,3:\\
1,\\
{x^2} + 4x + 3\\
= \left( {{x^2} + x} \right) + \left( {3x + 3} \right)\\
= x.\left( {x + 1} \right) + 3.\left( {x + 1} \right)\\
= \left( {x + 1} \right).\left( {x + 3} \right)\\
2,\\
{x^2} + 5x + 4\\
= \left( {{x^2} + x} \right) + \left( {4x + 4} \right)\\
= x.\left( {x + 1} \right) + 4.\left( {x + 1} \right)\\
= \left( {x + 1} \right).\left( {x + 4} \right)\\
3,\\
{x^2} - 4x + 3\\
= \left( {{x^2} - x} \right) + \left( { - 3x + 3} \right)\\
= x.\left( {x - 1} \right) - 3.\left( {x - 1} \right)\\
= \left( {x - 1} \right)\left( {x - 3} \right)\\
4,\\
{x^2} - 3x + 2\\
= \left( {{x^2} - x} \right) + \left( { - 2x + 2} \right)\\
= x\left( {x - 1} \right) - 2.\left( {x - 1} \right)\\
= \left( {x - 1} \right)\left( {x - 2} \right)\\
5,\\
{x^2} + 5x - 6\\
= \left( {{x^2} - x} \right) + \left( {6x - 6} \right)\\
= x\left( {x - 1} \right) + 6.\left( {x - 1} \right)\\
= \left( {x - 1} \right)\left( {x + 6} \right)\\
6,\\
{x^2} - 2x - 3\\
= \left( {{x^2} - 3x} \right) + \left( {x - 3} \right)\\
= x.\left( {x - 3} \right) + \left( {x - 3} \right)\\
= \left( {x - 3} \right)\left( {x + 1} \right)\\
7,\\
{x^2} + x - 6\\
= \left( {{x^2} - 2x} \right) + \left( {3x - 6} \right)\\
= x\left( {x - 2} \right) + 3.\left( {x - 2} \right)\\
= \left( {x - 2} \right)\left( {x + 3} \right)\\
8,\\
{x^2} - x - 6\\
= \left( {{x^2} - 3x} \right) + \left( {2x - 6} \right)\\
= x\left( {x - 3} \right) + 2.\left( {x - 3} \right)\\
= \left( {x - 3} \right)\left( {x + 2} \right)\\
9,\\
6{x^2} + 7x + 2\\
= \left( {6{x^2} + 4x} \right) + \left( {3x + 2} \right)\\
= 2x.\left( {3x + 2} \right) + \left( {3x + 2} \right)\\
= \left( {3x + 2} \right).\left( {2x + 1} \right)\\
10,\\
3{x^2} - 11x + 6\\
= \left( {3{x^2} - 9x} \right) + \left( { - 2x + 6} \right)\\
= 3x\left( {x - 3} \right) - 2.\left( {x - 3} \right)\\
= \left( {x - 3} \right)\left( {3x - 2} \right)
\end{array}\)
