Question

a)phân tích đa thức thành nhân tử:

\(x^4+2005x^2+2004x+2005\)

b)GTLN của:

\(-x^2-10y^2+6xy-2x+10y+9\)

Answer 1

a, ta có : \(x^4+2005x^2+2004x+2005\)

=\(x^4-x+2005x^2+2005x+2005\)

=\(x\left(x-1\right)\left(x^2+x+1\right)+2005\left(x^2+x+1\right)\)

=\(\left(x^2+x+1\right)\left(x^2-x+2005\right)\)

b, ta có \(-x^2-10y^2+6xy-2x+10y+9\)

=\(-\left(x^2+1+2x-6xy+9y^2-6y\right)-y^2+4y-4+13\)=\(13-\left(x-3y+1\right)^2-\left(y-2\right)^2\le13\forall x\)

Vậy Max=13 \(\Leftrightarrow\) \(\left\{{}\begin{matrix}x=5\\y=2\end{matrix}\right.\)