Ta có: \(3x^2+y^2-2xy+2y-10x+2029\)

\(=\left[\left(x^2-2xy+y^2\right)-\left(2x-2y\right)+1\right]+\left(2x^2-8x+8\right)+2020\)

\(=\left[\left(x-y\right)^2-2\left(x-y\right)+1\right]+2\left(x^2-4x+4\right)+2020\)

\(=\left(x-y-1\right)^2+2\left(x-2\right)^2+2020\ge2020\left(\forall x,y\right)\)

Dấu "=" xảy ra khi: \(\hept{\begin{cases}\left(x-y-1\right)^2=0\\2\left(x-2\right)^2=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=2\\y=1\end{cases}}\)

Vậy Min = 2020 khi x = 2 và y = 1

Ta có: \(a^4+a^2+1\)

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

\(=\left(a^2+1\right)^2-a^2\)

\(=\left(a^2-a+1\right)\left(a^2+a+1\right)\)

\(a^4+a^2+1=a^4+2a^2-a^2+1\)

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

\(=\left(a^2+1\right)^2-a^2\)

\(=\left(a^2+a+1\right)\left(a^2-a+1\right)\)

\(B=-2x^2+10x-8=-2x^2+10x-\frac{25}{2}+\frac{9}{2}\)

\(=-\left(2x^2-10x+\frac{25}{2}\right)+\frac{9}{2}\)

\(=-2\left(x^2-5x+\frac{25}{4}\right)+\frac{9}{2}\)

\(=-2\left[x^2-2.\frac{5}{2}x+\left(\frac{5}{2}\right)^2\right]+\frac{9}{2}\)

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

Vì \(\left(x-\frac{5}{2}\right)^2\ge0\forall x\)\(\Rightarrow-2\left(x-\frac{5}{2}\right)^2\le0\forall x\)

\(\Rightarrow-2\left(x-\frac{5}{2}\right)^2+\frac{9}{2}\le\frac{9}{2}\forall x\)

hay \(B\le\frac{9}{2}\)

Dấu " = " xảy ra \(\Leftrightarrow x-\frac{5}{2}=0\)\(\Leftrightarrow x=\frac{5}{2}\)

Vậy \(maxB=\frac{9}{2}\)\(\Leftrightarrow x=\frac{5}{2}\)

Ta có: 

\(B=-2x^2+10x-8\)

\(B=-2\left(x^2-5x+\frac{25}{4}\right)+\frac{9}{2}\)

\(B=-2\left(x-\frac{5}{2}\right)^2+\frac{9}{2}\le\frac{9}{2}\left(\forall x\right)\)

Dấu "=" xảy ra khi: x = 5/2

Vậy Max(B) = 9/2 khi x = 5/2

a, \(\frac{3x^3y^4}{-5xy^2}=-\frac{3}{5}x^2y^2\)

b, \(\frac{-4x^3y^5}{-\frac{1}{2}x^3y}=8y^4\)

c, \(\frac{5x^2y^2+10x^2y^2}{-5x^2y^3}=\frac{15x^2y^2}{-5x^2y^3}=\varnothing\)( Vì y^2  \(⋮̸̸\)y^3 )