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

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

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

\(x^2-y^2+yz-xz=\left(x-y\right)\left(x+y\right)-z\left(x-y\right)=\left(x-y\right)\left(x+y-z\right)\)

\(16x^2-\left(y^2+2y+1\right)=\left(4x\right)^2-\left(y+1\right)^2=\left(4x+y+1\right)\left(4x-y-1\right)\)

1+1=

\(=x^4-x^3+x^2-\left(x^3-x^2+x\right)+x^2-x+1\)

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

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

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

:3 lười giải 

\(6x^2+9x+12=6\left(x^2+2.\dfrac{3}{4}x+\dfrac{9}{16}\right)+\dfrac{69}{8}=6\left(x+\dfrac{3}{4}\right)^2+\dfrac{69}{8}\ge\dfrac{69}{8}\)

Dấu "=" xảy ra khi \(x=-\dfrac{3}{4}\)

a. \(2x\left(x+5\right)-3\left(x+5\right)=\left(x+5\right)\left(2x-3\right)\)

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