Đề thi online - Phân tích đa thức thành nhân tử bằng cách phối hợp nhiều phương pháp - Có lời giải chi tiết

Câu 1 : Phân tích đa thức thành nhân tử:\(\eqalign{& a)\;16{x^4}\left( {x - y} \right) - x + y \cr & c)\;16{x^3} - 54{y^3} \cr & e)\;{x^2} - 9 + \left( {2x + 7} \right)\left( {3 - x} \right) \cr & g)\;4{x^3} - 4{x^2} - x + 1 \cr} \) \(\eqalign{& b)\;2{x^3}y - 2x{y^3} - 4x{y^2} - 2xy \cr & d)\;{x^3} + {x^2} - 4x - 4 \cr & f)\;{x^2} - 2x + 1 - 4{y^2} \cr & h)\;{x^4} - 4{x^3} + 4{x^2} \cr} \)

A \(\eqalign{& a)\;\left( {2x - 1} \right)\left( {2x + 1} \right)\left( {4{x^2} + 1} \right)\left( {x - y} \right). \cr & b)\;2xy\left( {x - y - 1} \right)\left( {x + y + 1} \right). \cr & c)\;2\left( {2x - 3y} \right)\left( {4{x^2} + 6xy + 9{y^2}} \right). \cr & d)\;\left( {x - 2} \right)\left( {x + 2} \right)\left( {x + 1} \right). \cr & e)\left( {x - 3} \right)\left( { - x - 4} \right). \cr & f)\left( {x - 2y - 1} \right)\left( {x + 2y - 1} \right). \cr & g)\left( {2x - 1} \right)\left( {2x + 1} \right)\left( {x - 1} \right). \cr & h)\;{x^2}{\left( {x - 2} \right)^2}. \cr} \)

B \(\eqalign{& a)\;\left( {2x + 1} \right)\left( {2x + 1} \right)\left( {4{x^2} + 1} \right)\left( {x - y} \right). \cr & b)\;2xy\left( {x - y - 1} \right)\left( {x + y + 1} \right). \cr & c)\;2\left( {2x + 3y} \right)\left( {4{x^2} + 6xy + 9{y^2}} \right). \cr & d)\;\left( {x - 2} \right)\left( {x + 2} \right)\left( {x + 1} \right). \cr & e)\left( {x - 3} \right)\left( { - x - 4} \right). \cr & f)\left( {x - 2y - 1} \right)\left( {x + 2y - 1} \right). \cr & g)\left( {2x - 1} \right)\left( {2x + 1} \right)\left( {x - 1} \right). \cr & h)\;{x^2}{\left( {x - 2} \right)^2}. \cr} \)

C \(\eqalign{& a)\;\left( {2x - 1} \right)\left( {2x - 1} \right)\left( {4{x^2} + 1} \right)\left( {x + y} \right). \cr & b)\;2xy\left( {x - y - 1} \right)\left( {x + y + 1} \right). \cr & c)\;2\left( {2x + 3y} \right)\left( {4{x^2} + 6xy + 9{y^2}} \right). \cr & d)\;\left( {x - 2} \right)\left( {x + 2} \right)\left( {x + 1} \right). \cr & e)\left( {x - 3} \right)\left( { - x - 4} \right). \cr & f)\left( {x - 2y - 1} \right)\left( {x + 2y - 1} \right). \cr & g)\left( {2x - 1} \right)\left( {2x + 1} \right)\left( {x - 1} \right). \cr & h)\;{x^2}{\left( {x - 2} \right)^2}. \cr} \)

D \(\eqalign{& a)\;\left( {2x - 1} \right)\left( {2x + 1} \right)\left( {4{x^2} + 1} \right)\left( {x - y} \right). \cr & b)\;2xy\left( {x - y - 1} \right)\left( {x - y - 1} \right). \cr & c)\;2\left( {2x - 3y} \right)\left( {4{x^2} + 6xy + 9{y^2}} \right). \cr & d)\;\left( {x - 2} \right)\left( {x - 2} \right)\left( {x + 1} \right). \cr & e)\left( {x - 3} \right)\left( { - x - 4} \right). \cr & f)\left( {x - 2y - 1} \right)\left( {x + 2y - 1} \right). \cr & g)\left( {2x - 1} \right)\left( {2x + 1} \right)\left( {x - 1} \right). \cr & h)\;{x^2}{\left( {x - 2} \right)^2}. \cr} \)

Câu 2 : Phân tích đa thức thành nhân tử:\(a)\;{x^8} + {x^4} + 1\) \(b)\;{x^2} + 3x – 18\)

A \(\eqalign{& a)\,\,\left( {{x^3} - {x^3} + 1} \right)\left( {{x^2} - x + 1} \right)\left( {{x^2} + x + 1} \right). \cr & b)\,\left( {x + 6} \right)\left( {x - 3} \right). \cr} \)

B \(\eqalign{& a)\,\,\left( {{x^4} - {x^2} + 1} \right)\left( {{x^2} - x + 1} \right)\left( {{x^2} + x + 1} \right). \cr & b)\,\left( {x + 6} \right)\left( {x - 3} \right). \cr} \)

C \(\eqalign{& a)\,\,\left( {{x^4} - {x^2} + 1} \right)\left( {{x^2} - x + 1} \right)\left( {{x^2} + x + 1} \right). \cr & b)\,\left( {x - 6} \right)\left( {x + 3} \right). \cr} \)

D \(\eqalign{& a)\,\,\left( {{x^4} - {x^2} + 1} \right)\left( {{x^3} - x + 1} \right)\left( {{x^2} - x - 1} \right). \cr & b)\,\left( {x + 6} \right)\left( {x - 3} \right). \cr} \)

Câu 3 : Tìm x biết:\(\eqalign{& a)\;{\left( {2x - 3} \right)^2} - 4{x^2} + 9 = 0 \cr & c)\;2\left( {x + 3} \right) - {x^2} - 3x = 0 \cr} \) \(b)\;{x^4} + 2{x^3} - 8x - 16 = 0\)

A \(a)\,x = {3 \over 4},\,b)\,x = 2\,hoac\,x = 3,c)\,x = 2\,\,hoac\,\,x = - 2\)

B \(a)\,x = {5 \over 2},\,b)\,x = 2\,hoac\,x = - 3,c)\,x = 1\,\,hoac\,\,x = - 1\)

C \(a)\,x = {3 \over 2},\,b)\,x = 2\,hoac\,x = - 3,c)\,x = 2\,\,hoac\,\,x = - 2\)

D \(a)\,x = {3 \over 2},\,b)\,x = 4\,hoac\,x = - 4,c)\,x = 2\,\,hoac\,\,x = - 2\)

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