\(x^4+6x^2y+9y^2-1\)

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

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

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

\(=x^2\left(x-1\right)-\left(x-1\right)\)

\(=\left(x^2-1\right)\left(x-1\right)\)

\(=\left(x-1\right)\left(x+1\right)\left(x-1\right)\)

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

(x−1)2(x+1)

x² - 4x - y² + 4

= (x² - 4x + 4) - y²

= (x - 2)² - y²

= (x - y - 2)(x + y - 2)

(x - y - 2)(x + y - 2)

\(=x\left(a-b\right)-\left(a^2-2ab+b^2\right)=\)

\(=\left(a-b\right)x-\left(a-b\right)^2=\)

\(=\left(a-b\right)\left[x-\left(a-b\right)\right]=\left(a-b\right)\left(x-a+b\right)\)

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

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

\(=\left(x+1\right)\left(x-y\right)\)

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

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

\(=\left(x+1\right)\left(x-y\right)\)

Bạn xem lại đề

\(3\left(x+4\right)-x^2-4x\)

\(\Leftrightarrow3\left(x+4\right)-x\left(x+4\right)\)

\(\Leftrightarrow\left(3-x\right)\left(x+4\right)\)

\(=\left(x-y\right)\left(x+y\right)-10\left(x+y\right)=\)

\(=\left(x+y\right)\left(x-y-10\right)\)

= (x - y). (x + y) - 10 ( x - y)

= [( x + y) - 10)] . ( x - y)

  \(x^2\).(\(x\) - 1) + 16.( 1 - \(x\))

= \(x^2\).(\(x-1\)) - 16.( \(x-1\))

= (\(x\) - 1).(\(x^2\) - 16)

= (\(x\) - 1).(\(x\)² - 4²)

= (\(x\) - 1).(\(x\) - 4)(\(x\) + 4)