\(x^2+4xy+4y^2=x^2+2x.2y+\left(2y\right)^2=\left(x+2y\right)^2\)

\(x^2-2xy+y^2-36=\left(x-y\right)^2-6^2=\left(x-y+6\right)\left(x-y-6\right)\)

Ta có :

\(A=2x^2+9y^2-6xy-6x-12y+2004.\)

\(=\left(x^2-6xy+9y^2\right)+4\left(x-3y\right)+x^2-10x+2004\)

\(=\left(x-3y\right)^2=4\left(x-3y\right)+x^2-10y+25+1975\)

\(=\left(x-3y+2\right)^2+\left(x-5\right)^2+1975>1975\)

\(A_{min}=1975=x=5;y=\frac{7}{3}\)

Vậy

A = 2x²+9y²-6xy-6x-12y+2021

<=> A=x^2 + x^2 + 9y^2 - 6xy + 4x - 10x-12y + 1992 + 25 + 4 

<=> A=(x^2 - 6xy + 9y^2) + (4x-12y)+4+x^2-10x+25+1992

<=> A= (x- 3y)^2 + 4(x-3y) + 4 + (x-5)^2 +1992

<=> A = (x-3y+2)^2 + (x-5)^2 +1992

Vì : (x-3y+2)^2 + (x-5)^2  > 0

=>  (x-3y+2)^2 + (x-5)^2 +1992 > 1992

Dấu "=" xảy ra khi và chỉ khi :  \(\hept{\begin{cases}\left(x-5\right)^2=0\\\left(x-3y+2\right)^2=0\end{cases}\Leftrightarrow\hept{\begin{cases}x-5=0\\x-3y+2=0\end{cases}\Leftrightarrow}\hept{\begin{cases}x=5\\5-3y=-2\end{cases}}}\)

\(\hept{\begin{cases}x=5\\-3y=-2-5\end{cases}\Leftrightarrow\hept{\begin{cases}x=5\\-3y=-7\end{cases}}\Leftrightarrow\hept{\begin{cases}x=5\\y=\frac{7}{3}\end{cases}}}\)

Vậy Amin = 1992 khi x=5 ; y=7/3

hình như là xác định khi x=12 

Answer:

\(3x+2\left(5-x\right)=0\)

\(\Rightarrow3x+10-2x=0\)

\(\Rightarrow x+10=0\)

\(\Rightarrow x=-10\)

\(x\left(2x-1\right)\left(x+5\right)-\left(2x^2+1\right)\left(x+4,5\right)=3,5\)

\(\Rightarrow\left(2x^2-x\right)\left(x+5\right)-2x^3-x-9x^2-4,5=3,5\)

\(\Rightarrow2x^3-x^2+10x^2-5x-2x^3-x-9x^2=3,5+4,5\)

\(\Rightarrow\left(2x^3-2x^3\right)+\left(-x^2+10x^2-9x^2\right)+\left(-5x-x\right)=8\)

\(\Rightarrow-6x=8\)

\(\Rightarrow x=\frac{-4}{3}\)

ta có 

\(\left(x+1\right)\left(x-2\right)+5x\left(2-x\right)=\left(x+1\right)\left(x-2\right)-5x\left(x-2\right)\)

\(=\left(x-2\right)\left(x+1-5x\right)=0\Leftrightarrow\orbr{\begin{cases}x-2=0\\1-4x=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=2\\x=\frac{1}{4}\end{cases}}\)