Phân tích các đa thức sau thành nhân tử : a) x2 – 2x – 4y2 – 4y ; b) x4 + 2×3 – 4x – 4 ; 05/10/2021 Bởi Allison Phân tích các đa thức sau thành nhân tử : a) x2 – 2x – 4y2 – 4y ; b) x4 + 2×3 – 4x – 4 ;
\[\begin{array}{l} a)\,\,{x^2} – 2x – 4{y^2} – 4y\\ = \left( {{x^2} – 4{y^2}} \right) – 2\left( {x + 2y} \right)\\ = \left( {x – 2y} \right)\left( {x + 2y} \right) – 2\left( {x + 2y} \right)\\ = \left( {x + 2y} \right)\left( {x – 2y – 2} \right).\\ b)\,\,\,{x^4} + 2{x^3} – 4x – 4\\ = {x^4} – 4 + \left( {2{x^3} – 4x} \right)\\ = \left( {{x^2} – 2} \right)\left( {{x^2} + 2} \right) + 2x\left( {{x^2} – 2} \right)\\ = \left( {{x^2} – 2} \right)\left( {{x^2} + 2x + 2} \right). \end{array}\] Bình luận
a) \(x^2-2x-4y^2-4y\) \(=(x^2-4y^2)-2(x+2y)\) \(=(x+2y)(x-2y)-2(x+2y)\) \(=(x+2y)(x-2y-2)\). b) \( x^4+2x^3-4x-4\) \(=(x^4-4)+2x(x^2-2)\) \(=(x^2-2)(x^2+2)+2x(x^2-2)\) \(=(x^2-2)(x^2+2x+2)\). Bình luận
\[\begin{array}{l}
a)\,\,{x^2} – 2x – 4{y^2} – 4y\\
= \left( {{x^2} – 4{y^2}} \right) – 2\left( {x + 2y} \right)\\
= \left( {x – 2y} \right)\left( {x + 2y} \right) – 2\left( {x + 2y} \right)\\
= \left( {x + 2y} \right)\left( {x – 2y – 2} \right).\\
b)\,\,\,{x^4} + 2{x^3} – 4x – 4\\
= {x^4} – 4 + \left( {2{x^3} – 4x} \right)\\
= \left( {{x^2} – 2} \right)\left( {{x^2} + 2} \right) + 2x\left( {{x^2} – 2} \right)\\
= \left( {{x^2} – 2} \right)\left( {{x^2} + 2x + 2} \right).
\end{array}\]
a) \(x^2-2x-4y^2-4y\)
\(=(x^2-4y^2)-2(x+2y)\)
\(=(x+2y)(x-2y)-2(x+2y)\)
\(=(x+2y)(x-2y-2)\).
b) \( x^4+2x^3-4x-4\)
\(=(x^4-4)+2x(x^2-2)\)
\(=(x^2-2)(x^2+2)+2x(x^2-2)\)
\(=(x^2-2)(x^2+2x+2)\).