ủa tìm x sao ko có dấu =

x⁴ + 2005x² + 2004x + 2005 

=x⁴+2005x²+2005x-x+2005

=x⁴-x+2005x²+2005x+2005

=x(x³-1)+2005.(x²+x+1)

=x(x-1)(x²+x+1)+2005.(x²+x+1)

=(x²+x+1)[x(x-1)+2005]

=(x²+x+1)(x²-x+2005)

\(\frac{2004x}{2x^2+x+1}+\frac{2005x}{2x^2+x+1}=902\)

\(\Leftrightarrow\frac{2004x+2005x}{2x^2+x+1}=902\)

\(\Leftrightarrow\frac{4009x}{2x^2+x+1}=902\)

\(\Leftrightarrow4009x=902\left(2x^2+x+1\right)\)

\(\Leftrightarrow4009x=1804x^2+902x+902\)

\(\Leftrightarrow-1804x^2+3107x=902\)

Bn tự làm tiếp. Số to quá bn -.-

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

x⁴ + 2005x² + 2004x + 2005 

=x⁴-x+2005x²+2005x+2005

=x(x³-1)+2005.(x²+x+1)

=x(x-1)(x²+x+1)+2005.(x²+x+1)

=(x²+x+1)[x(x-1)+2005]

=(x²+x+1)(x²-x+2005)

x^4+2005x^2+2004x+2005

=x^4-x+2005x^2+2005x+2005

=x(x^3-1)+2005(x^2+x+1)

=x(x-1)(x^2+x+1)+2005(x^2+x+1)

=(x^2+x+1)(x^2-x+2005)