Đáp án:
\(\begin{array}{l}
1,\\
- 2.\left( {2x - 5} \right)\\
2,\\
\left( {x + 4} \right).\left( {3 - x} \right)\\
3,\\
5.\left( {x - y} \right)\left( {x + y - 2} \right)\\
4,\\
\left( {x - y} \right)\left( {x + 1} \right)\\
5,\\
\left( {a - b} \right).\left( {x - a + b} \right)\\
6,\\
\left( {x - y + 2} \right).\left( {x + y + 2} \right)\\
7,\\
{\left( {x - 1} \right)^2}.\left( {x + 1} \right)\\
8,\\
\left( {{x^2} + 3y - 1} \right).\left( {{x^2} + 3y + 1} \right)\\
9,\\
\left( {x + y} \right).\left( {x - 2} \right)\left( {x + 2} \right)\\
10,\\
\left( {x + 1} \right)\left( {{x^2} - 4x + 1} \right)\\
11,\\
3.\left( {x - y - 2z} \right).\left( {x - y + 2z} \right)\\
12,\\
\left( {x - 5} \right)\left( {x + 3} \right)\\
13,\\
\left( {x - 1} \right)\left( {2x + 5} \right)\\
14,\\
2.\left( {x - 3} \right)\left( {x + 3} \right)\\
15,\\
\left( {x - 2y} \right)\left( {x - 5y} \right)\\
16,\\
x.\left( {x - y - 1} \right)\left( {x + y - 1} \right)
\end{array}\)
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
1,\\
4{x^2} - 25 + \left( {2x + 7} \right)\left( {5 - 2x} \right)\\
= {\left( {2x} \right)^2} - {5^2} + \left( {2x + 7} \right).\left[ { - \left( {2x - 5} \right)} \right]\\
= \left( {2x - 5} \right).\left( {2x + 5} \right) - \left( {2x + 7} \right).\left( {2x - 5} \right)\\
= \left( {2x - 5} \right).\left[ {\left( {2x + 5} \right) - \left( {2x + 7} \right)} \right]\\
= \left( {2x - 5} \right).\left( {2x + 5 - 2x - 7} \right)\\
= \left( {2x - 5} \right).\left( { - 2} \right)\\
= - 2.\left( {2x - 5} \right)\\
2,\\
3.\left( {x + 4} \right) - {x^2} - 4x\\
= 3.\left( {x + 4} \right) - \left( {{x^2} + 4x} \right)\\
= 3.\left( {x + 4} \right) - x\left( {x + 4} \right)\\
= \left( {x + 4} \right).\left( {3 - x} \right)\\
3,\\
5{x^2} - 5{y^2} - 10x + 10y\\
= 5.\left( {{x^2} - {y^2} - 2x + 2y} \right)\\
= 5.\left[ {\left( {{x^2} - {y^2}} \right) + \left( { - 2x + 2y} \right)} \right]\\
= 5.\left[ {\left( {x - y} \right)\left( {x + y} \right) - 2.\left( {x - y} \right)} \right]\\
= 5.\left( {x - y} \right).\left[ {\left( {x + y} \right) - 2} \right]\\
= 5.\left( {x - y} \right)\left( {x + y - 2} \right)\\
4,\\
{x^2} - xy + x - y\\
= \left( {{x^2} - xy} \right) + \left( {x - y} \right)\\
= x.\left( {x - y} \right) + \left( {x - y} \right)\\
= \left( {x - y} \right)\left( {x + 1} \right)\\
5,\\
ax - bx - {a^2} + 2ab - {b^2}\\
= \left( {ax - bx} \right) - \left( {{a^2} - 2ab + {b^2}} \right)\\
= x.\left( {a - b} \right) - {\left( {a - b} \right)^2}\\
= \left( {a - b} \right).\left[ {x - \left( {a - b} \right)} \right]\\
= \left( {a - b} \right).\left( {x - a + b} \right)\\
6,\\
{x^2} + 4x - {y^2} + 4\\
= \left( {{x^2} + 4x + 4} \right) - {y^2}\\
= \left( {{x^2} + 2.x.2 + {2^2}} \right) - {y^2}\\
= {\left( {x + 2} \right)^2} - {y^2}\\
= \left[ {\left( {x + 2} \right) - y} \right].\left[ {\left( {x + 2} \right) + y} \right]\\
= \left( {x - y + 2} \right).\left( {x + y + 2} \right)\\
7,\\
{x^3} - {x^2} - x + 1\\
= \left( {{x^3} - {x^2}} \right) + \left( { - x + 1} \right)\\
= {x^2}.\left( {x - 1} \right) - \left( {x - 1} \right)\\
= \left( {x - 1} \right)\left( {{x^2} - 1} \right)\\
= \left( {x - 1} \right)\left( {{x^2} - {1^2}} \right)\\
= \left( {x - 1} \right).\left( {x - 1} \right)\left( {x + 1} \right)\\
= {\left( {x - 1} \right)^2}.\left( {x + 1} \right)\\
8,\\
{x^4} + 6{x^2}y + 9{y^2} - 1\\
= {\left( {{x^2}} \right)^2} + 2.{x^2}.3y + {\left( {3y} \right)^2} - 1\\
= {\left( {{x^2} + 3y} \right)^2} - {1^2}\\
= \left[ {\left( {{x^2} + 3y} \right) - 1} \right].\left[ {\left( {{x^2} + 3y} \right) + 1} \right]\\
= \left( {{x^2} + 3y - 1} \right).\left( {{x^2} + 3y + 1} \right)\\
9,\\
{x^3} + {x^2}y - 4x - 4y\\
= \left( {{x^3} + {x^2}y} \right) - \left( {4x + 4y} \right)\\
= {x^2}.\left( {x + y} \right) - 4.\left( {x + y} \right)\\
= \left( {x + y} \right)\left( {{x^2} - 4} \right)\\
= \left( {x + y} \right).\left( {{x^2} - {2^2}} \right)\\
= \left( {x + y} \right).\left( {x - 2} \right)\left( {x + 2} \right)\\
10,\\
{x^3} - 3{x^2} + 1 - 3x\\
= \left( {{x^3} + 1} \right) + \left( { - 3{x^2} - 3x} \right)\\
= \left( {{x^3} + {1^3}} \right) - 3.\left( {{x^2} + x} \right)\\
= \left( {x + 1} \right).\left( {{x^2} - x.1 + {1^2}} \right) - 3.x.\left( {x + 1} \right)\\
= \left( {x + 1} \right)\left( {{x^2} - x + 1} \right) - 3x.\left( {x + 1} \right)\\
= \left( {x + 1} \right).\left[ {\left( {{x^2} - x + 1} \right) - 3x} \right]\\
= \left( {x + 1} \right)\left( {{x^2} - 4x + 1} \right)\\
11,\\
3{x^2} - 6xy + 3{y^2} - 12{z^2}\\
= 3.\left( {{x^2} - 2xy + {y^2} - 4{z^2}} \right)\\
= 3.\left[ {\left( {{x^2} - 2xy + {y^2}} \right) - 4{z^2}} \right]\\
= 3.\left[ {{{\left( {x - y} \right)}^2} - {{\left( {2z} \right)}^2}} \right]\\
= 3.\left[ {\left( {x - y} \right) - 2z} \right].\left[ {\left( {x - y} \right) + 2z} \right]\\
= 3.\left( {x - y - 2z} \right).\left( {x - y + 2z} \right)\\
12,\\
{x^2} - 2x - 15\\
= \left( {{x^2} - 5x} \right) + \left( {3x - 15} \right)\\
= x.\left( {x - 5} \right) + 3.\left( {x - 5} \right)\\
= \left( {x - 5} \right)\left( {x + 3} \right)\\
13,\\
2{x^2} + 3x - 5\\
= \left( {2{x^2} - 2x} \right) + \left( {5x - 5} \right)\\
= 2x.\left( {x - 1} \right) + 5.\left( {x - 1} \right)\\
= \left( {x - 1} \right)\left( {2x + 5} \right)\\
14,\\
2{x^2} - 18\\
= 2.\left( {{x^2} - 9} \right)\\
= 2.\left( {{x^2} - {3^2}} \right)\\
= 2.\left( {x - 3} \right)\left( {x + 3} \right)\\
15,\\
{x^2} - 7xy + 10{y^2}\\
= \left( {{x^2} - 2xy} \right) + \left( { - 5xy + 10{y^2}} \right)\\
= x.\left( {x - 2y} \right) - 5y.\left( {x - 2y} \right)\\
= \left( {x - 2y} \right)\left( {x - 5y} \right)\\
16,\\
{x^3} - 2{x^2} + x - x{y^2}\\
= x.\left( {{x^2} - 2x + 1 - {y^2}} \right)\\
= x.\left[ {\left( {{x^2} - 2x + 1} \right) - {y^2}} \right]\\
= x.\left[ {\left( {{x^2} - 2.x.1 + {1^2}} \right) - {y^2}} \right]\\
= x.\left[ {{{\left( {x - 1} \right)}^2} - {y^2}} \right]\\
= x.\left[ {\left( {x - 1} \right) - y} \right].\left[ {\left( {x - 1} \right) + y} \right]\\
= x.\left( {x - y - 1} \right)\left( {x + y - 1} \right)
\end{array}\)
