Bài 9. Phân tích đa thức thành nhân tử

\(2{x^2} + x = 0 \Leftrightarrow x\left( {2x + 1} \right) = 0 \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{x = 0}\\{2x + 1 = 0}\end{array}} \right. \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{x = 0}\\{x = \dfrac{{ - 1}}{2}}\end{array}} \right.\)

Vậy \(x = 0;x = \dfrac{{ - 1}}{2}\)

\({x^2} - 2xy = x.x - 2xy = x\left( {x - 2y} \right)\)

a, \(6y^3+2y=2y\left(3y^3+1\right).\)

b, \(4\left(x-y\right)-3x\left(x-y\right)=\left(4-3x\right)\left(x-y\right).\)

\(2{x^2} + x = 0 \Leftrightarrow x\left( {2x + 1} \right) = 0 \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{x = 0}\\{2x + 1 = 0}\end{array}} \right. \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{x = 0}\\{x = \dfrac{{ - 1}}{2}}\end{array}} \right.\)

Vậy \(x = 0;x = \dfrac{{ - 1}}{2}\)

a) \({\left( {x + 1} \right)^2} - {y^2} = \left( {x + 1 + y} \right)\left( {x + 1 - y} \right)\)

b) \({x^3} + 3{x^2} + 3x + 1 = {\left( {x + 1} \right)^3}\)

c) \(8{x^3} - 12{x^2} + 6x - 1 = {\left( {2x} \right)^3} - 3.{\left( {2x} \right)^2}.1 + 3.2x.1 - {1^3} = {\left( {2x - 1} \right)^3}\)

\(2{x^2} - 4xy + 2y - x = \left( {2{x^2} - 4xy} \right) + \left( {2y - x} \right) = 2x\left( {x - 2y} \right) - \left( {x - 2y} \right) = \left( {x - 2y} \right)\left( {2x - 1} \right)\)

\(A = {x^2} + 2y - 2x - xy \\= \left( {{x^2} - 2x} \right) + \left( {2y - xy} \right) \\= x\left( {x - 2} \right) + y\left( {2 - x} \right) \\= x\left( {x - 2} \right) - y\left( {x - 2} \right) \\= (x-y)(x-2) \)

Thay \(x = 2022,y = 2020\) vào A ta được:

\(A = (2022 - 2020)(2022-2) = 2.2020 = 4040\)

\({x^3} - x = x\left( {{x^2} - 1} \right) = x\left( {x - 1} \right)\left( {x + 1} \right)\)

Bạn Tròn có kết quả đúng, bạn Vuông chưa phân tích triệt để.

\(\begin{array}{l}a)\,{x^2} + xy = x.x + x.y = x\left( {x + y} \right); \\b)\,6{a^2}b - 18ab = 6ab\left( {a - 3} \right); \\c)\,{x^3} - 4x = x\left( {{x^2} - 4} \right) = x\left( {x - 2} \right)\left( {x + 2} \right); \\d)\,{x^4} - 8x = x\left( {{x^3} - 8} \right) = x\left( {x - 2} \right)\left( {{x^2} + 2x + 4} \right).\end{array}\)

a)     

\({x^2} - 9 + xy + 3y \\= \left( {{x^2} - 9} \right) + \left( {xy + 3y} \right) \\= \left( {x - 3} \right)\left( {x + 3} \right) + y\left( {x + 3} \right) \\= \left( {x + 3} \right)\left( {x - 3 + y} \right)\)

b)     

\({x^2}y + {x^2} + xy - 1 \\= \left( {{x^2}y + xy} \right) + \left( {{x^2} - 1} \right) \\= xy\left( {x + 1} \right) + \left( {x + 1} \right)\left( {x - 1} \right) \\= \left( {x + 1} \right)\left( {xy + x - 1} \right)\)