Question

sử dụng biện pháp thêm bớt

x⁴+2002x² +2001x+2002

Answer 1

\(x^4+2002x^2+2001x+2002\)

\(=x^4+x^3-x^3+x^2-x^2+2002x^2+2002x-x+2002\)

\(=\left(x^4+x^3+x^2\right)-\left(x^3+x^2+x\right)+\left(2002x^2+2002x+2002\right)\)

\(=x^2\left(x^2+x+1\right)-x\left(x^2+x+1\right)+2002\left(x^2+x+1\right)\)

\(=\left(x^2+x+1\right)\left(x^2-x+2002\right)\)

Answer 2

Ta có: \(x^4+2002x^2+2001x+2002\)

= \(x^4+2002x^2+2002x-x+2002\)

= \(\left(x^4-x\right)+2002\left(x^2+x+1\right)\)

= \(x\left(x^3-1\right)+2002\left(x^2+x+1\right)\)

= \(x\left(x-1\right)\left(x^2+x+1\right)+2002\left(x^2+x+1\right)\)

= \(\left(x^2+x+1\right)\left[x\left(x-1\right)+2002\right]\)

=\(\left(x^2+x+1\right)\left(x^2-x+2002\right)\)