tick cho mình rồi mình làm cho

x^4-4x-1

= x^4 + 4 + 4x² - 4x² = ( x² + 2 )² - 4x² = ( x² + 2 - 2x )( x² + 2 + 2x )

hoặc

 x^4-4x-1

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

3{x-2}²/x{x-2}{x²+2x+4}=3{x-2}/x²+2x+4=-3/x+2 

nho tick minh nha

\(A=\frac{5x^2+4x-1}{x^2}=\frac{9x^2-\left(4x^2-4x+1\right)}{x^2}=9-\frac{\left(2x-1\right)^2}{x^2}\le9\)

Dấu \(=\)khi \(2x-1=0\Leftrightarrow x=\frac{1}{2}\).

\(B=\frac{x^2}{x^2+x+1}=\frac{3x^2}{3x^2+3x+3}=\frac{4x^2+4x+4-\left(x^2+4x+4\right)}{3x^2+3x+3}=\frac{4}{3}-\frac{\left(x+2\right)^2}{3\left(x^2+x+1\right)}\le\frac{4}{3}\)

Dấu \(=\)khi \(x+2=0\Leftrightarrow x=-2\).

dùng hằng đẳng thúc cho mẫu rút gọn ta được 
\(\frac{1}{x^2+x+1}-\frac{1}{Y^2+y+1}+\frac{2\left(x+y\right)}{x^2y^2+3}\)=\(\frac{y^2+y+1-x^2-x-1}{\left(x^2+x+1\right)\left(y^2+y+1\right)}+\frac{2\left(x-y\right)}{x^2y^2+3}\)
=\(\frac{\left(y-x\right)\left(y+x\right)+\left(y-x\right)}{x^2y^2+x^2y+x^2+xy^2+xy+x+y^2+y+1}+\frac{2\left(x-y\right)}{x^2y^2+3}\)
=\(\frac{-2\left(x-y\right)}{xy\left(x+y\right)+\left(x+y\right)+1+x^2y^2+x^2+y^2+xy}+\frac{2\left(x-y\right)}{x^2y^2+3}\)
=\(\frac{-2\left(x-y\right)}{2xy+x^2+y^2+x^2y^2+2}+\frac{2\left(x-y\right)}{x^2y^2+3}\)