Question

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Bài 1. Phân tích đa thức thành nhân tử bằng phương pháp đặt nhân tử chung

a/   ( x + 2 )²  +  2( x² - 4 ) + ( x - 2 )²

b/    2x² - 2xy - 4y² 

c/    x² - 2x - 4y² - 4y

d/    4x( x - 2y ) - 8y( x - 2y )

Answer 1

[Nguyễn Minh Đăng]

a) \(\left(x+2\right)^2+2\left(x^2-4\right)+\left(x-2\right)^2\)

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

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

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

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

\(=2x\cdot2x=4x^2\)

b) \(2x^2-2xy-4y^2\)

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

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

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

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

Answer 2

[Nguyễn Minh Đăng]

c) \(x^2-2x-4y^2-4y\)

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

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

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

d) \(4x\left(x-2y\right)-8y\left(x-2y\right)\)

\(=\left(x-2y\right)\left(4x-8y\right)\)

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

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