Phân tích đa thức thành nhân tử a) xy - 16(x^2)y b) (x^2) - {y^2} - 3x + 3y c) (x^2)- 4xy + 4(y^2) - 4
1.
a | \(xy - 16{x^2}y = xy\left( {1 - 16x} \right)\) |
b | \({x^2} - {y^2} - 3x + 3y\) \( = \left( {{x^2} - {y^2}} \right) - 3\left( {x - y} \right)\) \[ = \left( {x - y} \right)\left( {x + y} \right) - 3\left( {x - y} \right)\]\[ = \left( {x - y} \right)\left( {x + y - 3} \right)\] |
c | \({x^2} - 4xy + 4{y^2} - 4\)\({\rm{ = }}\left( {{x^2} - 4xy + 4{y^2}} \right) - 4\) \({\rm{ = }}{\left( {x - 2y} \right)^2} - {2^2}\) \( = \left( {x - 2y - 2} \right)\left( {x - 2y + 2} \right)\) |
2.
a | \(15{x^2} - 3x = 0\) \(3x\left( {5x - 1} \right) = 0\) TH1: \(x = 0\) TH2: \[5x - 1 = 0\] \[x = \frac{1}{5}\] Vậy \[x = \frac{1}{5}\] |
b | \[2x\left( {x - 7} \right) + 4\left( {7 - x} \right) = 0\] \[2x\left( {x - 7} \right) - 4\left( {x - 7} \right) = 0\] \[\left( {x - 7} \right)\left( {2x - 4} \right) = 0\] |
| TH1: \[\begin{array}{l}x - 7 = 0\\x = 7\end{array}\] ] TH2: \[\begin{array}{l}2x - 4 = 0\\x = 2\end{array}\] Vậy \[x \in \left\{ {7\,;\,\,5} \right\}.\] |
c | \(x\left( {2x + 3} \right) + 4{x^2} - 9 = 0\) \(x\left( {2x + 3} \right) + \left( {2x + 3} \right)\left( {2x - 3} \right) = 0\) \(\left( {2x + 3} \right)\left( {x + 2x - 3} \right) = 0\) \(\left( {2x + 3} \right)\left( {3x - 3} \right) = 0\) |
| TH1: \(2x + 3 = 0\) \(x = \frac{{ - 3}}{2}\) TH2: \(3x - 3 = 0\) \(x = 1\) Vậy \(x \in \left\{ {\frac{{ - 3}}{2};\,\,1} \right\}.\) |