Toán Lớp 8: Bài 2. (*)Phân tích các đa thức sau thành nhân tử.
1. ????3 − 4????2 + 12???? − 27
2. ????3 + 2????2 + 2???? + 1
3. ????4 − 2????3 + 2???? − 1
4. ????4 + 2????3 + 2????2 + 2???? + 1
5. ????2 − 2???? − 4????2 − 4????
6. ????4 + 2????3 − 4???? − 4
Bài 3. Phân tích các đa thức sau thành nhân tử.
1. ????2 + 4???? + 3
2. ????2 + 5???? + 4
3. ????2 − 4???? + 3
4. ????2 − 3???? + 2
5. ????2 + 5???? − 6
6. ????2 − 2???? − 3
7. ????2 + ???? − 6
8. ????2 − ???? − 6
9. 6????2 + 7???? + 2
10. 3????2 − 11???? + 6
11. 2????2 + 5???? − 3
12. ???? + 2????2 − 6
13. 7???? − 6????2 − 2
14. 16???? − 5????2 − 3
15. −7????2 + 11???? + 6
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Comments ( 2 )
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}\)
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}\)