\(N=100+180y+81y^2+9y^2\)

\(=90y^2+180y+100=10\left(9y^2+18y+10\right)\)\(=10\left[9\left(y^2+2y+1\right)+1\right]\)

\(=10\left[9\left(y+1\right)^2+1\right]=90\left(y+1\right)^2+10\)

vì \(90\left(y+1\right)^2\ge0\Rightarrow90\left(y+1\right)^2+10\ge10\)

vậy GTNN của N là 10 khi \(y+1=0\Leftrightarrow y=-1\)

B = x² + 9y² + 2011

= x² + (3y)² + 2011

= x² + (5 - 2x)² + 2011 (do 2x + 3y = 5)

= x² + 4x² - 20x + 25 + 2011

= 5x² - 20x + 2036

= 5x² - 20x + 20 + 2016

= 5(x² - 4x + 4)  + 2016

= 5(x - 2)² + 2016 \(\ge2016\)

=> Min B = 2016 khi x - 2 = 0 <=> x = 2 

khi đó y = \(\dfrac{1}{3}\)

Vậy B_min = 2016 khi x = 2 ; \(y=\dfrac{1}{3}\)

H=\(x^6-2x^3+x^2-2x+2\)

\(=x^6+2x^5+3x^4+2x^2-2x^5-4x^4-6x^3-4x^2-4x+x^4+2x^3+3x^2+2x+2\)

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

\(=\left(x^2-2x+1\right)\left(x^4+2x^3+3x^2+2\right)\)

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

\(=\left(x-1\right)^2\left(x^2+1\right)\left[\left(x+1\right)^2+1\right]\text{≥}0\)

Vì \(\left\{{}\begin{matrix}\left(x-1\right)^2\text{≥}0\\\left(x^2+1\right)\text{≥}1\\\left(x+1\right)^2+1\text{≥}1\end{matrix}\right.\)

⇒ MinH=0 ⇔ \(x=1\)

\(A=5x^2+9y^2-12xy+24x-48y+81\)

\(A=4x^2+x^2+9y^2-12xy+32x-48y-8x+16+1+64\)

\(A=(4x^2+9y^2+64-12xy+32x-48y)+\left(x^2-8x+16\right)+1\)

\(A=[\left(2x\right)^2+\left(3y\right)^2+\left(8\right)^2-2.2x.3y-2.3y.8+2.2x.8]+\left(x^2-8x+16\right)+1\)

\(A=\left(2x-3y+8\right)^2\left(x-4\right)^2+1\)

\(Do\) \(\left(2x-3y+8\right)^2\ge0\) \(và\) \(\left(x-4\right)^2\ge0\)

\(\Rightarrow A_{min}=1\)

anh ko biết nha em yêu của anh

\(A=2x^2+9y^2-6xy-6x-12y+2046\)

\(=\left[\left(x^2-6xy+9y^2\right)+\left(4x-12y\right)+4\right]-4+\left(x^2-10x+25\right)-25+2046\)

\(=\left[\left(x-3y\right)^2+4\left(x-3y\right)+4\right]+\left(x-5\right)^2-4-25+2046\)

\(=\left(x-3y+2\right)^2+\left(x-5\right)^2+2017\ge2017\)

Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}x-3y+2=0\\x-5=0\end{cases}\Leftrightarrow\hept{\begin{cases}y=\frac{7}{3}\\x=5\end{cases}}}\)

Vậy \(A_{min}=2017\) tại \(x=5;y=\frac{7}{3}\)