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 \(\begin{array}{l}1)\,\,2{y^2}\left( {xy - 2} \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,3)\,\,\left( {x + y} \right)\left( {x - y + 1} \right)\\2)\,\,y{\left( {x - 3} \right)^2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,4)\,\,\left( {x - 3} \right)\left( {x - 1} \right)\end{array}\)
Câu hỏi trên thuộc đề trắc nghiệm
Đề thi HK1 Toán 8 - Trường Tạ Quang Bửu - Hà Nội - Năm 2017 - 2018 (có giải chi tiết).