Đề đúng: \(C=x^2+4y^2+2x-4y-4xy+2011\)

\(C=\left(x^2-4xy+4y^2\right)+\left(2x-4y\right)+1+2010\)

\(C=\left(x-2y\right)^2+2\left(x-2y\right)+1+2010\)

\(C=\left(x-2y+1\right)^2+2010\ge2010\)

Dấu "=" xảy ra khi: \(\left(x-2y+1\right)^2=0\)

=> Ta có vô số cặp (x;y) thỏa mãn ví dụ như:

(1;1) ; (-1;0) ; (3;2) ; ...

C = x² + 4y² + 2x - 4y - 4xy + 2011 ( đúng chưa :v )

C = [ ( x² - 4xy + 4y² ) + 2x - 4y + 1 ] + 2010

C = [ ( x - 2y )² + 2( x - 2y ) + 1 ] + 2010

C = [ ( x - 2y ) + 1 ]² + 2010

C = ( x - 2y + 1 )² + 2010 ≥ 2010 ∀ x,y 

Đẳng thức xảy ra <=> x - 2y + 1 = 0

                            <=> x - 2y = -1

                            <=> x = 2y - 1

=> MinC = 2011 <=> x = 2y - 1

 Ta có: 
A=x^2 - 2*3x + 9 +2(y^2 - 2y +1) + 7 
=(x-3)^2 +2(y-1)^2 +7 >+ 7 
=> minA= 7 <=> x=3 và y=1

C = (x-1)² + ( y -2)² +7 -4 -1

GTNN C = 3

a) \(A=\left(x-1\right)\left(x+2\right)\left(x+3\right)\left(x+6\right)\)

\(A=\left[\left(x-1\right)\left(x+6\right)\right]\left[\left(x+2\right)\left(x+3\right)\right]\)

\(A=\left(x^2+5x-6\right)\left(x^2+5x+6\right)\)

\(A=\left(x^2+5x\right)^2-6^2\)

\(A=\left(x^2+5x\right)^2-36\)

Vì \(\left(x^2+5x\right)^2\ge0\) với mọi x

\(\Rightarrow\left(x^2+5x\right)^2-36\ge-36\)

\(\Rightarrow Amin=-36\Leftrightarrow x^2+5x=0\)

\(\Rightarrow x\left(x+5\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x=-5\end{matrix}\right.\)

b) \(B=x^2-2x+y^2+4y+8\)

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

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

Vì \(\left(x-1\right)^2\ge0\) với mọi x

\(\left(y+2\right)^2\ge0\) với mọi y

\(\Rightarrow\left(x-1\right)^2+\left(y+2\right)^2\ge0\) với mọi x,y

\(\Rightarrow\left(x-1\right)^2+\left(y+2\right)^2+3\ge3\)

\(\Rightarrow Bmin=3\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-2\end{matrix}\right.\)

c) \(C=x^2-4x+y^2-8y+6\)

\(C=\left(x^2-4x+4\right)+\left(y^2-8y+16\right)-14\)

\(C=\left(x-2\right)^2+\left(y-4\right)^2-14\)

Vì \(\left(x-2\right)^2\ge0\) với mọi x

\(\left(y-4\right)^2\ge0\) với mọi y

\(\Rightarrow\left(x-2\right)^2+\left(y-4\right)^2-14\ge-14\) với mọi x,y

\(\Rightarrow Cmin=-14\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=4\end{matrix}\right.\)

\(H=x^2-2x+y^2-4y+7=\left(x^2-2x+1\right)+\left(y^2-4y+4\right)+2=\left(x-1\right)^2+\left(y-2\right)^2+2\ge2\)

\(minH=2\Leftrightarrow\)\(\left\{{}\begin{matrix}x=1\\y=2\end{matrix}\right.\)

\(C=x^2+2x-y^2+4y-7\)

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

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

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

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

\(\Rightarrow C=\left(x+1\right)^2-\left(y-2\right)^2-4\ge0-0-4\)

\(\Rightarrow C\ge-4\)

Vậy\(GTNN_C=-4\)tại \(x=-1\)và  \(y=2\)

1)

Ta có:

\(A=x^2+5y^2-2xy+4y+3=(x^2+y^2-2xy)+4y^2+4y+3\)

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

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

Thấy rằng: \((x-y)^2\geq 0; (2y+1)^2\geq 0 , \forall x,y\)

\(\Rightarrow A\geq 0+0+2=2\)

Vậy GTNN của $A$ là $2$. Dấu "=" xảy ra khi \(\left\{\begin{matrix} (x-y)^2=0\\ (2y+1)^2=0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x=-\frac{1}{2}\\ y=\frac{1}{2}\end{matrix}\right.\)

2)

Đặt \(x^2-2x=a\)

Khi đó: \(B=a(a+2)=a^2+2a+1-1=(a+1)^2-1\)

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

\(=(x-1)^4-1\)

Thấy rằng \((x-1)^4\geq 0, \forall x\Rightarrow B\geq 0-1=-1\)

Vậy GTNN của $B$ là $-1$ khi \((x-1)^4=0\Leftrightarrow x=1\)

A=x²+5y²-2xy+4y+32

=(x^2-2xy+4y²)+(y²+4y+4)+28

=(x-2y)²+(y+2)²+28

Vì (x-2y)²+(y+2)² lớn hơn hoặc bằng 0 với mọi x,y thuộc R

=> (x-2y)²+(y+2)²+28 lớn hơn hoặc bằng 28 với mọi x,y thuộc R

Hay A lớn hơn hoặc bằng 28=> GTNN của A là 28

A=28<=>

+) y+2=0=> y=-2

+) x-2.(-2)=0 => x=-4

A=x²+2xy+2y²-2x-4y+2

=x²+xy-x+y²+xy-y-x-y+1+y²-2y+1

=(x²+xy-x)+(y²+xy-y)-(x+y-1)+(y²-2y+1)

= x(x+y-1)+y(y+x-1)-(x+y-1)+(y-1)²

=(x+y-1)(x+y-1)+(y-1)²

A=(x+y-1)²+(y-1)²

do (x+y-1)²\(\ge0\forall x;y\)

(y-1)²\(\ge0\forall y\)

=>(x+y-1)²+(y-1)²\(\ge0\)

=>Min A=0 khi

x+y-1=0

=>x+y=1 (*)

y-1=0

=>y=1

thay y=1 vào (*) ta đc

x+1=1

=>x=0

vậy....

3) \(B=3x^2+x+7\)

\(\Leftrightarrow B=3x^2+x+\dfrac{1}{12}+\dfrac{83}{12}\)

\(\Leftrightarrow B=3\left(x^2+\dfrac{1}{3}x+\dfrac{1}{36}\right)+\dfrac{83}{12}\)

\(\Leftrightarrow B=3\left[x^2+2.x.\dfrac{1}{6}+\left(\dfrac{1}{6}\right)^2\right]+\dfrac{83}{12}\)

\(\Leftrightarrow B=3\left(x+\dfrac{1}{6}\right)^2+\dfrac{83}{12}\)

Vậy GTNN của \(B=\dfrac{83}{12}\) khi \(x+\dfrac{1}{6}=0\Leftrightarrow x=\dfrac{-1}{6}\)

B=3x² +x+7

=3x2+x+\(\dfrac{1}{12}+\dfrac{83}{12}\)

=3\(\left(x^2+\dfrac{1}{3}x+\dfrac{1}{36}\right)+\dfrac{83}{12}\)

=3 \(\left(x+\dfrac{1}{6}\right)^2+\dfrac{83}{12}\)

do \(\left(x+\dfrac{1}{6}\right)^2\ge0\forall x\)

=> \(3\left(x+\dfrac{1}{6}\right)^2\ge0\)

=> 3\(\left(x+\dfrac{1}{6}\right)^2+\dfrac{83}{12}\ge\dfrac{83}{12}\)

GTNN B=\(\dfrac{83}{12}\)

khi x+\(\dfrac{1}{6}\) =0

=>x=-\(\dfrac{1}{6}\)

=x² +4y²+4xy+1+2x+4y+x²+9

=(x+2y)²+2(x+2y)+1+x²+9

=(x+2y+1)²+x²+9

có (x+2y+1)²≥0 với mọi x,y

x²≥0 với mọi x

⇒(x+2y+1)²+x² ≥0với mọi x,y

⇒(x+2y+1)²+x²+9≥9với mọi x,y

⇒

ta có :

A = 2x²+4y²+4xy+2x+4y+9 = 2x²+2x+4y²+4y+4xy+9

= 2x(x+1)+4y(y+1)+4xy+9

= 2x(x+1)+4y(y+x+1)+9

= (x+1)(2x+4y²)+9

=> A lớn hơn hoặc bằng 9

=> min A là 9

\(2x^2+4y^2+4xy+2x+4y+9\\ =x^2+x^2+4y^2+4xy+2x+4y+1+8\\ =\left(x^2+4xy+4y^2\right)+\left(2x+4y\right)+1+x^2+8\\ =\left(x+2y\right)^2+2\left(x+2y\right)+1+x^2+8\\ =\left[\left(x+2y\right)^2+2\left(x+2y\right)+1\right]+x^2+8\\ =\left(x+2y+1\right)^2+x^2+8\\ Do\text{ }x^2\ge0\forall x\\ \left(x+2y+1\right)^2\ge0\forall x;y\\ \Rightarrow\left(x+2y+1\right)^2+x^2\ge0\forall x;y\\ \Rightarrow\left(x+2y+1\right)^2+x^2+8\ge8\forall x;y\\ Dấu\text{ }"="\text{ }xảy\text{ }ra\text{ }khi:\left\{{}\begin{matrix}x^2=0\\\left(x+2y+1\right)^2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\x+2y+1=0\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=0\\2y+0+1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\2y=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\y=-\dfrac{1}{2}\end{matrix}\right.\\ Vậy\text{ }GTNN\text{ }của\text{ }biểu\text{ }thức\text{ }là\text{ }8\text{ }khi\text{ }\left\{{}\begin{matrix}x=0\\y=-\dfrac{1}{2}\end{matrix}\right.\)