Bài 2:

a: Sửa đề: \(A=-4x^2-5y^2+8xy+10y+12\)

\(=-4x^2+8xy-4y^2-y^2+10y-25+37\)

\(=-\left(2x-2y\right)^2-\left(y-5\right)^2+37\le37\forall x,y\)

Dấu '=' xảy ra khi \(\begin{cases}2x-2y=0\\ y-5=0\end{cases}\Rightarrow\begin{cases}y=5\\ x=y=5\end{cases}\)

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

\(=-\frac14\left(4x^2+4y^2-4xy-8x-8y\right)\)

\(=-\frac14\left(4x^2-4xy+y^2-8x+4y+3y^2-12y\right)\)

\(=-\frac14\left\lbrack\left(2x-y\right)^2-4\left(2x-y\right)+4+3y^2-12y+12-16\right\rbrack\)

\(=-\frac14\left\lbrack\left(2x-y-2\right)^2+3\left(y-2\right)^2-16\right\rbrack=-\frac14\left(2x-y-2\right)^2-\frac34\left(y-2\right)^2+4\le4\forall x,y\)

Dấu '=' xảy ra khi y-2=0 và 2x-y-2=0

=>y=2 và 2x=y+2=2+2=4

=>x=2 và y=2

Bài 1:

d: \(D=2x^2+3y^2+4xy-8x-2y\)

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

\(=2\left(x+y\right)^2-8\left(x+y\right)+8+y^2+6y+9-17\)

\(=2\left(x+y-2\right)^2+\left(y+3\right)^2-17\ge-17\forall x,y\)

Dấu '=' xảy ra khi \(\begin{cases}y+3=0\\ x+y-2=0\end{cases}\Rightarrow\begin{cases}y=-3\\ x=-y+2=-\left(-3\right)+2=3+2=5\end{cases}\)

f: \(F=2x^2+8xy+11y^2-4x-2y+6\)

\(=2x^2+8xy+8y^2-4x-8y+3y^2+6y+6\)

\(=2\left(x+2y\right)^2-4\left(x+2y\right)+2+3y^2+6y+3+1\)

\(=2\left(x+2y-1\right)^2+3\left(y+1\right)^2+1\ge1\forall x,y\)

Dấu '=' xảy ra khi y+1=0 và x+2y-1=0

=>y=-1 và x=-2y+1=-2*(-1)+1=2+1=3

h: \(H=x^2+y^2-xy-x+y+1\)

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

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

\(=\frac14\left\lbrack\left(2x-y\right)^2-2\left(2x-y\right)+1+3y^2+2y+\frac13+\frac83\right\rbrack\)

\(=\frac14\cdot\left\lbrack\left(2x-y-1\right)^2+3\left(y+\frac13\right)^2+\frac83\right\rbrack\ge\frac23\forall x,y\)

Dấu '=' xảy ra khi \(\begin{cases}y+\frac13=0\\ 2x-y-1=0\end{cases}\Rightarrow\begin{cases}y=-\frac13\\ 2x=y+1=-\frac13+1=\frac23\end{cases}\Rightarrow\begin{cases}y=-\frac13\\ x=\frac13\end{cases}\)

Bài 2:

a: Sửa đề: \(A=-4x^2-5y^2+8xy+10y+12\)

\(=-4x^2+8xy-4y^2-y^2+10y-25+37\)

\(=-\left(2x-2y\right)^2-\left(y-5\right)^2+37\le37\forall x,y\)

Dấu '=' xảy ra khi \(\begin{cases}2x-2y=0\\ y-5=0\end{cases}\Rightarrow\begin{cases}y=5\\ x=y=5\end{cases}\)

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

\(=-\frac14\left(4x^2+4y^2-4xy-8x-8y\right)\)

\(=-\frac14\left(4x^2-4xy+y^2-8x+4y+3y^2-12y\right)\)

\(=-\frac14\left\lbrack\left(2x-y\right)^2-4\left(2x-y\right)+4+3y^2-12y+12-16\right\rbrack\)

\(=-\frac14\left\lbrack\left(2x-y-2\right)^2+3\left(y-2\right)^2-16\right\rbrack=-\frac14\left(2x-y-2\right)^2-\frac34\left(y-2\right)^2+4\le4\forall x,y\)

Dấu '=' xảy ra khi y-2=0 và 2x-y-2=0

=>y=2 và 2x=y+2=2+2=4

=>x=2 và y=2

Bài 1:

d: \(D=2x^2+3y^2+4xy-8x-2y\)

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

\(=2\left(x+y\right)^2-8\left(x+y\right)+8+y^2+6y+9-17\)

\(=2\left(x+y-2\right)^2+\left(y+3\right)^2-17\ge-17\forall x,y\)

Dấu '=' xảy ra khi \(\begin{cases}y+3=0\\ x+y-2=0\end{cases}\Rightarrow\begin{cases}y=-3\\ x=-y+2=-\left(-3\right)+2=3+2=5\end{cases}\)

f: \(F=2x^2+8xy+11y^2-4x-2y+6\)

\(=2x^2+8xy+8y^2-4x-8y+3y^2+6y+6\)

\(=2\left(x+2y\right)^2-4\left(x+2y\right)+2+3y^2+6y+3+1\)

\(=2\left(x+2y-1\right)^2+3\left(y+1\right)^2+1\ge1\forall x,y\)

Dấu '=' xảy ra khi y+1=0 và x+2y-1=0

=>y=-1 và x=-2y+1=-2*(-1)+1=2+1=3

h: \(H=x^2+y^2-xy-x+y+1\)

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

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

\(=\frac14\left\lbrack\left(2x-y\right)^2-2\left(2x-y\right)+1+3y^2+2y+\frac13+\frac83\right\rbrack\)

\(=\frac14\cdot\left\lbrack\left(2x-y-1\right)^2+3\left(y+\frac13\right)^2+\frac83\right\rbrack\ge\frac23\forall x,y\)

Dấu '=' xảy ra khi \(\begin{cases}y+\frac13=0\\ 2x-y-1=0\end{cases}\Rightarrow\begin{cases}y=-\frac13\\ 2x=y+1=-\frac13+1=\frac23\end{cases}\Rightarrow\begin{cases}y=-\frac13\\ x=\frac13\end{cases}\)

Ta có:

D=2x2+3y2+4xy−8x−2y+18C=2x2+3y2+4xy−8x−2y+18

D=2(x2+2xy+y2)+y2−8x−2y+18C=2(x2+2xy+y2)+y2−8x−2y+18

D=2[(x+y)2−4(x+y)+4]+(y2+6y+9)+1C=2[(x+y)2−4(x+y)+4]+(y2+6y+9)+1

D=2(x+y−2)2+(y+3)2+1≥1C=2(x+y−2)2+(y+3)2+1≥1

Dấu "=" xảy ra ⇔x+y=2⇔x+y=2và y=−3y=−3

Hay x = 5 , y = -3

Đc chx bạn

\(-x^2-y^2+xy+2x+2y=-\left[x^2-x\left(y+2\right)+\dfrac{1}{4}\left(y+2\right)^2\right]-\left(\dfrac{3}{4}y^2-3y+3\right)+4=-\left(x-\dfrac{1}{2}y-1\right)^2-\left(\dfrac{\sqrt{3}}{2}y-\sqrt{3}\right)^2+4\le4\)

\(max=4\Leftrightarrow\)\(\left\{{}\begin{matrix}x=2\\y=2\end{matrix}\right.\)

Thanks

Ta có

\(B=\dfrac{xy^2+y^2\left(y^2-x\right)+2}{x^2y^4+y^4+2x^2+2}\)

\(B=\dfrac{xy^2+y^4-xy^2+2}{y^4\left(x^2+1\right)+2\left(x^2+1\right)}\)

\(B=\dfrac{y^4+2}{\left(x^2+1\right)\left(y^4+2\right)}\)

B=\(\dfrac{1}{x^2+1}\)

Ta có:

x^2\(\ge0\)

x²+1\(\ge1\)

\(\dfrac{1}{x^2+1}\le1\)

\(\Rightarrow B\le1\)

Dấu "=" xảy ra khi

x²=0

=>x=0

Vậy GTLN của B là 1 khi x=0

(2x-y)² >=0 với mọi x,y. 

<=> 4x²+y²>=4xy với mọi x,y. 

<=> 4x²+y²+4xy>=4xy+4xy với mọi x,y. 

<=> 8xy<=(2x+y)² với mọi x,y.

=> 8xy<=36 vì 2x+y=6.

=> xy<=4,5 hay A<=4,5.

Dấu bằng sảy ra khi 2x-y=0 <=> x=y/2 thay vào 2x+y=6 ta được x=3/2,y=3.

a, A=3.(2/3)^3-2.(1/2)^3-6.(2/3)^2.(1/2)^2+(2/3).(1/2)

      =8/9-1/4-2/3+1/3=8/9-1/4-1/3=11/36

b,  B=-1+(-1/18)+1/12+2/3=-11/36

\(2x+2y+z=4\Rightarrow z=4-2x-2y\)

Ta có: \(A=2xy+yz+xz\)

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

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

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

  \(\Rightarrow2A=8y-4xy-4y^2+8x-4x^2\)

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

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

               \(=-\left[4x^2+2.x.2\left(y-2\right)+\left(y-2\right)^2\right]+\left(y-2\right)^2-4y^2+8y\)

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

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

                     \(=-\left(2x+y-2\right)^2-3\left(y^2-2.\frac{2}{3}y+\frac{4}{9}-\frac{4}{9}-\frac{4}{3}\right)\)       

                      \(=-\left(2x+y-2\right)^2-3\left(y-\frac{2}{3}\right)^2+\frac{16}{3}\)

                        \(=\frac{16}{3}-\left[\left(2x+y-2\right)^2+3\left(y-\frac{2}{3}\right)^2\right]\)

Vì \(\left(2x+y-2\right)^2\ge0;\left(y-\frac{2}{3}\right)^2\ge0\) Nên \(\frac{16}{3}-\left[\left(2x+y-2\right)^2+3\left(y-\frac{2}{3}\right)^2\right]\le\frac{16}{3}\)

\(\Rightarrow A\le\frac{16}{3}:2=\frac{8}{3}\)

Dấu "=" xảy ra <=>\(\hept{\begin{cases}y-\frac{2}{3}=0\\2x+y-2=0\\z=4-2x-2y\end{cases}}\Leftrightarrow\hept{\begin{cases}x=\frac{-y+2}{2}\\y=\frac{2}{3}\\z=4-2x-2y\end{cases}\Leftrightarrow\hept{\begin{cases}x=\frac{2}{3}\\y=\frac{2}{3}\\z=\frac{4}{3}\end{cases}}}\)

Vậy A_Max = 8/3 khi và chỉ khi x = y = 2/3 và z = 4/3