Bài 1 (6 điểm): Phân tích các đa thức sau thành nhân tử: a) (a + b)2 – m2 + a + b – m b) x3 - 6x2 + 12x – 8 c) x2 – 7xy + 10y2 d) x4 + 2x3 - 4x – 4 Bài 2 (4 điểm): Tìm x, biết: a) (3x – 5)2 – (x +1 )2 = 0 b) (5x – 4)2 – 49x2 = 0

Đáp án:

$\begin{array}{l}
B1)\\
a){\left( {a + b} \right)^2} - {m^2} + a + b - m\\
 = \left( {a + b + m} \right)\left( {a + b - m} \right) + \left( {a + b - m} \right)\\
 = \left( {a + b - m} \right)\left( {a + b + m + 1} \right)\\
b){x^3} - 6{x^2} + 12x - 8\\
 = {x^3} - 3.{x^2}.2 + 3.x{.2^2} - {2^3}\\
 = {\left( {x - 2} \right)^3}\\
c){x^2} - 7xy + 10{y^2}\\
 = {x^2} - 2xy - 5xy + 10{y^2}\\
 = x\left( {x - 2y} \right) - 5y\left( {x - 2y} \right)\\
 = \left( {x - 2y} \right)\left( {x - 5y} \right)\\
d){x^4} + 2{x^3} - 4x - 4\\
 = \left( {{x^4} - 4} \right) + \left( {2{x^3} - 4x} \right)\\
 = \left( {{x^2} - 2} \right)\left( {{x^2} + 2} \right) + 2x\left( {{x^2} - 2} \right)\\
 = \left( {{x^2} - 2} \right)\left( {{x^2} + 2 + 2x} \right)\\
B2)\\
a){\left( {3x - 5} \right)^2} - {\left( {x + 1} \right)^2} = 0\\
 \Rightarrow \left( {3x - 5 - x - 1} \right)\left( {3x - 5 + x + 1} \right) = 0\\
 \Rightarrow \left( {2x - 6} \right)\left( {4x - 4} \right) = 0\\
 \Rightarrow \left[ \begin{array}{l}
2x - 6 = 0\\
4x - 4 = 0
\end{array} \right.\\
 \Rightarrow \left[ \begin{array}{l}
x = 3\\
x = 1
\end{array} \right.\\
b){\left( {5x - 4} \right)^2} - 49{x^2} = 0\\
 \Rightarrow {\left( {5x - 4} \right)^2} - {\left( {7x} \right)^2} = 0\\
 \Rightarrow \left( {5x - 4 - 7x} \right)\left( {5x - 4 + 7x} \right) = 0\\
 \Rightarrow \left( { - 2x - 4} \right)\left( {12x - 4} \right) = 0\\
 \Rightarrow \left[ \begin{array}{l}
x =  - 2\\
x = \dfrac{1}{3}
\end{array} \right.
\end{array}$