a)  \(x^4+2x^3+5x^2+4x-12\)

\(=x^4+x^3+6x^2+x^3+x^2+6x-2x^2-2x-12\)

\(=x^2\left(x^2+x+6\right)+x\left(x^2+x+6\right)-2\left(x^2+x+6\right)\)

\(=\left(x^2+x+6\right)\left(x^2+x-2\right)\)

\(=\left(x^2+x+6\right)\left(x-1\right)\left(x+2\right)\)

Đặt \(a+b-2c=x,b+c-2a=y,c+a-2b=z\)

\(\Rightarrow x+y+z=0\)

Chắc bạn biết: \(x+y+z=0\Rightarrow x^3+y^3+z^3=3xyz\)

Vậy \(\left(a+b-2c\right)^3+\left(b+c-2a\right)^3+\left(c+a-2b\right)^3=3\left(a+b-2c\right)\left(b+c-2a\right)\left(c+a-2b\right)\)

Chúc bạn học tốt.

Đặt:  \(a+b-2c=x;\)   \(b+c-2a=y;\)\(c+a-2b=z\)

=>   \(x+y+z=0\)

=>  \(x^3+y^3+z^3=3xyz\)

Thay trở lại ta được:

\(\left(a+b-2c\right)^3+\left(b+c-2a\right)^3+\left(c+a-2b\right)^3\)

\(=3\left(a+b-2c\right)\left(b+c-2a\right)\left(c+a-2b\right)\)

1) \(ab\left(a+b\right)-bc\left(b+c\right)+ac\left(a-c\right)\)

\(=ab\left(a+b\right)-b^2c-bc^2+a^2c-ac^2\)

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

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

\(=\left(a+b\right)\left(ab-ac+bc-c^2\right)\)

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

\(=\left(a+b\right)\left(b-c\right)\left(a+c\right)\)

a)  \(x^2-8x+16-y^2=\left(x-4\right)^2-y^2=\left(x-y-4\right)\left(x+y-4\right)\)

b)  \(4x^2-y^2+10y-25=4x^2-\left(y-5\right)^2=\left(2x-y+5\right)\left(2x+y-5\right)\)

c)  \(x^2-y^2+z^2-t^2-2xz+2yt=\left(x-z\right)^2-\left(y-t\right)^2=\left(x-y-z+t\right)\left(x+y-x-t\right)\)

p/s: chúc bạn học tốt

 (x-1)^2-2(x-1)(2y-1)+(2y-1) = [ x-1 - (2y-1)]² = ( x-1-2y+1)² = ( x-2y)²

mk chỉnh lại đề

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

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

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

p/s: chúc bạn học tốt