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Đáp án: $\begin{array}{l}
6)\left( {x + 5} \right)\left( {x - 3} \right)\\
7)\left( {x + y} \right)\left( {x + y + 1} \right)\left( {x + y - 1} \right)\\
8)\left( {x - 3} \right)\left( {x - 2} \right)\left( {x + 2} \right)\\
9)\left( {x - 2y} \right)\left( {x - 5y} \right)\\
10)\left( {{a^2} + 2 + 2a} \right)\left( {{a^2} + 2 - 2a} \right)\\
11)x\left( {2x + 5} \right)\left( {x - 1} \right)\\
12)\left( {y + z} \right)\left( {3{x^2} + 3xy + yz + 3xz} \right)\\
13)2.\left( {x - 2} \right)\left( {{x^2} + 1} \right)\\
14)\left( {x + 2} \right)\left( {{x^2} - 6x + 4} \right)\\
15)\left( {x - 4} \right)\left( {3x + 2} \right)\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)\\
16)\left( {x - 2y + 3z} \right)\left( {x + 4y - z} \right)
\end{array}$
Giải thích các bước giải:
$\begin{array}{l}
6){x^2} + 2x - 15\\
= {x^2} + 5x - 3x - 15\\
= x\left( {x + 5} \right) - 3\left( {x + 5} \right)\\
= \left( {x + 5} \right)\left( {x - 3} \right)\\
7){x^3} - x + 3{x^2}y + 3x{y^2} - y + {y^3}\\
= \left( {{x^3} + 3{x^2}y + 3x{y^2} + {y^3}} \right) - \left( {x + y} \right)\\
= {\left( {x + y} \right)^3} - \left( {x + y} \right)\\
= \left( {x + y} \right)\left( {{{\left( {x + y} \right)}^2} - 1} \right)\\
= \left( {x + y} \right)\left( {x + y + 1} \right)\left( {x + y - 1} \right)\\
8){x^3} - 3{x^2} - 4x + 12\\
= {x^2}\left( {x - 3} \right) - 4\left( {x - 3} \right)\\
= \left( {x - 3} \right)\left( {{x^2} - 4} \right)\\
= \left( {x - 3} \right)\left( {x - 2} \right)\left( {x + 2} \right)\\
9){x^2} - 7xy + 10{y^2}\\
= {x^2} - 2xy - 5xy + 10{y^2}\\
= x\left( {x - 2y} \right) - 5y\left( {x - 2y} \right)\\
= \left( {x - 2y} \right)\left( {x - 5y} \right)\\
10){a^4} + 4\\
= {a^4} + 4{a^2} + 4 - 4{a^2}\\
= \left( {{a^2} + 2} \right) - {\left( {2a} \right)^2}\\
= \left( {{a^2} + 2 + 2a} \right)\left( {{a^2} + 2 - 2a} \right)\\
11)2{x^3} + 3{x^2} - 5x\\
= x\left( {2{x^2} + 3x - 5} \right)\\
= x\left( {2{x^2} + 5x - 2x - 5} \right)\\
= x\left( {2x + 5} \right)\left( {x - 1} \right)\\
12){\left( {x + y + z} \right)^3} - {x^3} - {y^3} - {z^3}\\
= \left( {x + y + z - x} \right)\left[ {{{\left( {x + y + z} \right)}^2} + x.\left( {x + y + z} \right) + {x^2}} \right]\\
- \left( {y + z} \right)\left( {{y^2} + yz + {z^2}} \right)\\
= \left( {y + z} \right)\left( {{x^2} + {y^2} + {z^2} + 2xy + 2xz + 2yz + {x^2} + xy + xz + {x^2}} \right)\\
- \left( {y + z} \right)\left( {{y^2} + yz + {z^2}} \right)\\
= \left( {y + z} \right)\left( {3{x^2} + {y^2} + {z^2} + 3xy + 2yz + 3xz - {y^2} - yz - {z^2}} \right)\\
= \left( {y + z} \right)\left( {3{x^2} + 3xy + yz + 3xz} \right)\\
13)\\
2{x^3} - 4{x^2} + 2x - 4\\
= 2{x^2}\left( {x - 2} \right) + 2\left( {x - 2} \right)\\
= 2.\left( {x - 2} \right)\left( {{x^2} + 1} \right)\\
14)\\
{x^3} - 4{x^2} - 8x + 8\\
= {x^3} + 8 - 4{x^2} - 8x\\
= \left( {x + 2} \right)\left( {{x^2} - 2x + 4} \right) - 4x\left( {x + 2} \right)\\
= \left( {x + 2} \right)\left( {{x^2} - 2x + 4 - 4x} \right)\\
= \left( {x + 2} \right)\left( {{x^2} - 6x + 4} \right)\\
15)3{x^5} - 10{x^4} - 8{x^3} - 3{x^2} + 10x + 8\\
= {x^3}\left( {3{x^2} - 10x - 8} \right) - \left( {3{x^2} - 10x - 8} \right)\\
= \left( {3{x^2} - 10x - 8} \right)\left( {{x^3} - 1} \right)\\
= \left( {3{x^2} - 12x + 2x - 8} \right).\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)\\
= \left( {x - 4} \right)\left( {3x + 2} \right)\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)\\
16)\\
{x^2} + 2xy - 8{y^2} + 2xz + 14yz - 3{z^2}\\
= \left( {{x^2} + {y^2} + {z^2} + 2xy + 2xz + 2yz} \right)\\
+ \left( { - 9{y^2} + 12yz - 4{z^2}} \right)\\
= {\left( {x + y + z} \right)^2} - \left( {9{y^2} - 12yz + 4{z^2}} \right)\\
= {\left( {x + y + z} \right)^2} - {\left( {3y - 2z} \right)^2}\\
= \left( {x + y + z - 3y + 2z} \right)\left( {x + y + z + 3y - 2z} \right)\\
= \left( {x - 2y + 3z} \right)\left( {x + 4y - z} \right)
\end{array}$
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