Toán Lớp 8: Giúp mik với ạ!!! Cần gấp!
Bài 1: Phân tích đa thức thành nhân tử:
1. x^2 – 5x – 6
2. x^2 + x -2
3. 3x^2 – 2x – 5
4. 9x^2 + 25x – 6
5. x^2 – 7x + 6
6. x^2 + 8x +7
7. x^2 x – 6
8. x^2 + 4x – 5
9. x^2 + 4x -21
10. 2x^2 + 3x -2
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Comments ( 1 )
1,\\
\left( {x – 6} \right).\left( {x + 1} \right)\\
2,\\
\left( {x – 1} \right).\left( {x + 2} \right)\\
3,\\
\left( {3x – 5} \right).\left( {x + 1} \right)\\
4,\\
\left( {9x – 2} \right)\left( {x + 3} \right)\\
5,\\
\left( {x – 1} \right)\left( {x – 6} \right)\\
6,\\
\left( {x + 1} \right).\left( {x + 7} \right)\\
7,\\
\left( {x – 3} \right)\left( {x + 2} \right)\\
8,\\
\left( {x – 1} \right)\left( {x + 5} \right)\\
9,\\
\left( {x – 3} \right).\left( {x + 7} \right)\\
10,\\
\left( {2x – 1} \right).\left( {x + 2} \right)
\end{array}\)
1,\\
{x^2} – 5x – 6\\
= \left( {{x^2} – 6x} \right) + \left( {x – 6} \right)\\
= x.\left( {x – 6} \right) + \left( {x – 6} \right)\\
= \left( {x – 6} \right).\left( {x + 1} \right)\\
2,\\
{x^2} + x – 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)\\
3,\\
3{x^2} – 2x – 5\\
= \left( {3{x^2} – 5x} \right) + \left( {3x – 5} \right)\\
= x.\left( {3x – 5} \right) + \left( {3x – 5} \right)\\
= \left( {3x – 5} \right).\left( {x + 1} \right)\\
4,\\
9{x^2} + 25x – 6\\
= \left( {9{x^2} – 2x} \right) + \left( {27x – 6} \right)\\
= x.\left( {9x – 2} \right) + 3.\left( {9x – 2} \right)\\
= \left( {9x – 2} \right)\left( {x + 3} \right)\\
5,\\
{x^2} – 7x + 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} + 8x + 7\\
= \left( {{x^2} + x} \right) + \left( {7x + 7} \right)\\
= x.\left( {x + 1} \right) + 7.\left( {x + 1} \right)\\
= \left( {x + 1} \right).\left( {x + 7} \right)\\
7,\\
{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)\\
8,\\
{x^2} + 4x – 5\\
= \left( {{x^2} – x} \right) + \left( {5x – 5} \right)\\
= x.\left( {x – 1} \right) + 5.\left( {x – 1} \right)\\
= \left( {x – 1} \right)\left( {x + 5} \right)\\
9,\\
{x^2} + 4x – 21\\
= \left( {{x^2} – 3x} \right) + \left( {7x – 21} \right)\\
= x.\left( {x – 3} \right) + 7.\left( {x – 3} \right)\\
= \left( {x – 3} \right).\left( {x + 7} \right)\\
10,\\
2{x^2} + 3x – 2\\
= \left( {2{x^2} – x} \right) + \left( {4x – 2} \right)\\
= x.\left( {2x – 1} \right) + 2.\left( {2x – 1} \right)\\
= \left( {2x – 1} \right).\left( {x + 2} \right)
\end{array}\)