Question

[help nhanh]

Bài 1:Phân tích đa thức thành nhân tử (5 điểm)

a) 4x + 6y

b) 2x(x+y) + 3(x+y)

c) xy(x-3) – 2(3-x)

d) x2 – 4x+4 – y2

e) x2 + 4x – 2xy - 4y + y2

Answer 1

[Yen Nhi]

\(4x+6y=2.\left(2x+3y\right)\)

\(2x.\left(x+y\right)+3.\left(x+y\right)=\left(2x+3\right).\left(x+y\right)\)

\(xy.\left(x-3\right)-2.\left(3-x\right)=-xy.\left(3-x\right)-2.\left(3-x\right)=\left(3-x\right).\left(-xy-2\right)\)

\(x^2-4x+4-y^2\)

\(=\left(x^2-4x+4\right)-y^2\)

\(=\left(x-2\right)^2-y^2\)

\(=\left(x-2-y\right).\left(x-2+y\right)\)

\(x^2+4x-2xy-4y+y^2\)

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

\(=\left(x-y\right)^2+4.\left(x-y\right)\)

\(=\left(x-y\right).(x-y+4)\)