Phân tích đa thức thành nhân tử:1) \(2x{y^2} - 4y\...

Câu hỏi: Phân tích đa thức thành nhân tử:1) \(2x{y^2} - 4y\)2) \({x^2}y - 6xy + 9y\)3) \({x^2} + x - {y^2} + y\)4) \({x^2} + 4x + 3\)

A \(\begin{array}{l}1)\,\,2y\left( {xy - 2} \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,3)\,\,\left( {x + y} \right)\left( {x - y + 1} \right)\\2)\,\,y{\left( {x - 3} \right)^2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,4)\,\,\left( {x + 3} \right)\left( {x + 1} \right)\end{array}\)

B \(\begin{array}{l}1)\,\,2{y^2}\left( {xy - 2} \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,3)\,\,\left( {x + y} \right)\left( {x - y + 1} \right)\\2)\,\,y{\left( {x + 3} \right)^2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,4)\,\,\left( {x + 3} \right)\left( {x + 1} \right)\end{array}\)

C \(\begin{array}{l}1)\,\,2y\left( {xy - 2} \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,3)\,\,\left( {x - y} \right)\left( {x + y + 1} \right)\\2)\,\,y{\left( {x + 3} \right)^2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,4)\,\,\left( {x + 3} \right)\left( {x + 1} \right)\end{array}\)

D \(\begin{array}{l}1)\,\,2{y^2}\left( {xy - 2} \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,3)\,\,\left( {x + y} \right)\left( {x - y + 1} \right)\\2)\,\,y{\left( {x - 3} \right)^2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,4)\,\,\left( {x - 3} \right)\left( {x - 1} \right)\end{array}\)