\(Q=x^2+2y^2-2xy-4y+2017\)

\(Q=\left(x^2-2xy+y^2\right)+\left(y^2-4y+4\right)+2013\)

\(Q=\left(x-y\right)^2+\left(y-2\right)^2+2013\ge2013\)

Vậy GTNN của Q=2013 <=> \(\orbr{\begin{cases}x-y=0\\y-2=0\end{cases}}\)<=>\(\orbr{\begin{cases}\\\end{cases}}x=y=2\)

b: \(B=x^2+xy+y^2-3x-3y\)

\(=\frac14\left(4x^2+4xy+4y^2-12x-12y\right)\)

\(=\frac14\left(4x^2+4xy+y^2-12x-6y+3y^2-6y\right)\)

\(=\frac14\left\lbrack\left(2x+y\right)^2-6\left(2x+y\right)+9+3y^2-6y+3-12\right\rbrack\)

\(=\frac14\left\lbrack\left(2x+y-3\right)^2+3\left(y-1\right)^2-12\right\rbrack\ge\frac14\cdot\left(-12\right)=-3\forall x,y\)

Dấu '=' xảy ra khi y-1=0 và 2x+y-3=0

=>y=1 và 2x=-y+3=-1+3=2

=>y=1 và x=1

a: \(A=x^2-2xy+2y^2+2x-10y+17\)

\(=x^2-2xy+y^2+2x-2y+y^2-8y+17\)

\(=\left(x-y\right)^2+2\left(x-y\right)+1+y^2-8y+16\)

\(=\left(x-y+1\right)^2+\left(y-4\right)^2\ge0\forall x,y\)

Dấu '=' xảy ra khi y-4=0 và x-y+1=0

=>y=4 và x=y-1=4-1=3

c: \(C=2x^2+9y^2-6xy-6x-12y+2008\)

\(=x^2-6xy+9y^2+4x-12y+x^2-10x+2008\)

\(=\left(x-3y\right)^2+4\left(x-3y\right)+4+x^2-10x+25+1979\)

\(=\left(x-3y+2\right)^2+\left(x-5\right)^2+1979\ge1979\forall x,y\)

Dấu '=' xảy ra khi x-5=0 và x-3y+2=0

=>x=5 và 3y=x+2=5+2=7

=>x=5 và y=7/3

Ta có \(C=x^2+2y^2-2xy-4y+5=\left(x-2xy+y^2\right)+\left(y^2-4y+4\right)+1\)

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

Do \(\left(x-y\right)^2\ge0;\left(y-2\right)^2\ge0\Rightarrow C\ge1\)

Vậy GTNN của C là 1 khi \(\hept{\begin{cases}x-y=0\\y-2=0\end{cases}}\Leftrightarrow x=y=2\)

2) a) \(P=3x^2+y^2-8x+2xy+16\)

\(P=\left(x^2+2xy+y^2\right)+2\left(x^2-4x+4\right)+8\)

\(P=\left(x+y\right)^2+2\left(x-2\right)^2+8\ge8\forall x;y\)

\(\Rightarrow\) GTNN của P là 8 khi \(\left\{{}\begin{matrix}\left(x+y\right)^2=0\\\left(x-2\right)^2=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x+y=0\\x-2=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}y=-x\\x=2\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=-2\end{matrix}\right.\) vậy GTNN của P là 8 khi \(x=2;y=-2\)

b) \(Q=x^2+2y^2-2xy-4y+2017\)

\(Q=\left(x^2-2xy+y^2\right)+\left(y^2-4y+4\right)+2013\)

\(Q=\left(x-y\right)^2+\left(y-2\right)^2+2013\ge2013\forall x;y\)

\(\Rightarrow\) GTNN của Q là 2013 khi \(\left\{{}\begin{matrix}\left(x-y\right)^2=0\\\left(y-2\right)^2=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x-y=0\\y-2=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=y\\y=2\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}y=2\\x=2\end{matrix}\right.\) vậy GTNN của Q là 2013 khi \(x=y=2\)

c) \(M=2x^2+y^2-2xy-2x+2016\)

\(M=\left(x^2-2xy+y^2\right)+\left(x^2-2x+1\right)+2015\)

\(M=\left(x-y\right)^2+\left(x-1\right)^2+2015\ge2015\forall x;y\)

\(\Rightarrow\) GTNN của M là 2015 khi \(\left\{{}\begin{matrix}\left(x-y\right)^2=0\\\left(x-1\right)^2=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x-y=0\\x-1=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=y\\x=1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=1\end{matrix}\right.\) vậy GTNN của M là 2015 khi \(x=y=1\)

|3x-7|+|3x-2|+8 >= 5+8 = 13 

Dấu "=" xảy ra <=> 3/2 <= x <= 7/3

k mk nha

tiếp đi bạn 

\(D=2x^2+2xy+5y^2-8x-22y\)

\(=2x^2+2x(y-4)+(5y^2-22y)\)

\(=2\left\lbrack x^2+2x\cdot\frac{y-4}{2}+\left(\frac{y-4}{2}\right)^2\right\rbrack-2\left(\frac{y-4}{2}\right)^2+5y^2-22y\)

\(=2\left(x+\frac{y-4}{2}\right)^2-\frac{y^2-8y+16}{2}+5y^2-22y\)

\(=2\left(x+\frac{y-4}{2}\right)^2+\frac92y^2-18y-8\)

\(=2\left(x+\frac{y-4}{2}\right)^2+\frac92\left(y^2-4y+4\right)-26\)

\(=2\left(x+\frac{y-4}{2}\right)^2+\frac92\left(y-2\right)^2-26\ge-26\forall x,y\)

Dấu '=' xảy ra khi y-2=0 và \(x+\frac{y-4}{2}=0\)

=>y=2 và \(x=\frac{-y+4}{2}=\frac{-2+4}{2}=\frac22=1\)

\(N=2x^2+y^2+2xy-4x-2y+3\)

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

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

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

Mà  \(\left(x+y-1\right)\ge0\forall x;y\)

       \(\left(x-1\right)^2\ge0\forall x\)

\(\Rightarrow N\ge1\)

Dấu "=" xảy ra khi :

\(\hept{\begin{cases}x+y-1=0\\x-1=0\end{cases}}\Leftrightarrow\hept{\begin{cases}y=0\\x=1\end{cases}}\)

Vậy  \(N_{Min}=1\Leftrightarrow\left(x;y\right)=\left(1;0\right)\)

\(N=2x^2+y^2+2xy-4x-2y\)\(+3\)

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

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

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

\(Do\)\(\left(x+y+1\right)^2\)\(\ge\)\(0\)\(\forall\)\(x\)\(;\)\(y\)

\(x^2\)\(\ge\)\(0\)\(\forall\)\(x\)

=.>\(\left(x+y+1\right)^2+x^2+2\)\(\ge\)\(2\)\(\forall\)\(x\)\(;\)\(y\)

=>\(N\)\(\ge\)\(2\)\(\forall\)\(x\)\(;\)\(y\)

Dấu = xảy ra khi: 

\(\hept{\begin{cases}\left(x+y+1\right)^2=0\\x^2=0\end{cases}}\)

=>\(\hept{\begin{cases}x+y+1=0\\x=0\end{cases}}\)

=>\(\hept{\begin{cases}x+y=-1\\x=0\end{cases}}\)

=>\(\hept{\begin{cases}y=-1\\x=0\end{cases}}\)

Vậy \(N_{min}\)\(=\)\(2\)khi \(y=-1\)\(;\)\(x=0\)

Chúc pạn họk tốt~~~!!! :3

\(A=2x^2+2xy+y^2-2x+2y+1\)

\(A=x^2+2xy+y^2+2x+2y+x^2-4x+4+1-4\)

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

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

Vì \(\left(x+y+1\right)^2\ge0\forall x;y\)và \(\left(x-2\right)^2\ge0\forall x\)

\(\Rightarrow A\ge-4\forall x;y\)

Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}x+y+1=0\\x-2=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=2\\y=-3\end{cases}}}\)

Vậy....

\(P=3x^2+y^2-8x+2xy+16\)

\(P=\left(x^2+2xy+y^2\right)+\left(2x^2-8x+8\right)+8\)

\(P=\left(x+y\right)^2+2\left(x-2\right)^2+8\ge8\)

Vậy GTNN của P=8 <=> \(\orbr{\begin{cases}x+y=0\\x-2=0\end{cases}}\)<=>\(\orbr{\begin{cases}y=-2\\x=2\end{cases}}\)