Bài 1:

$xy+3=x+y$

$\Leftrightarrow xy-x-y+3=0$

$\Leftrightarrow x(y-1)-(y-1)+2=0$

$\Leftrightarrow (x-1)(y-1)+2=0$
$\Leftrightarrow (x-1)(y-1)=-2$
Vì $x,y$ nguyên nên $x-1, y-1$ nguyên. Khi đó:

$(x-1, y-1)=(2, -1), (-2, 1), (1, -2), (-1, 2)$
Đến đây bạn dễ dàng tìm được giá trị $x,y$ thỏa mãn.

Bài 2:

$x+y=3\Rightarrow y=3-x$. Khi đó:

$A=xy=x(3-x)=3x-x^2$

$-A=x^2-3x=(x^2-3x+1,5^2)-1,5^2=(x-1,5)^2-\frac{9}{4}\geq \frac{-9}{4}$

$\Rightarrow A\leq \frac{9}{4}$

Vậy $A_{\max}=\frac{9}{4}$ 

 Ta có :

\(x^2+y^2+xy=3\)

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

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

hay \(S^2=3+xy\le3+\frac{\left(x+y\right)^2}{4}=3+\frac{S^2}{4}\)

\(\Rightarrow S^2\le3+\frac{S^2}{4}\)

\(\Rightarrow S^2\le4\)

\(\Rightarrow-2\le S\le2\)

GTLN của S = 2

\(3=\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{xy}\Leftrightarrow x+y+1=3xy\)

\(\Leftrightarrow y\left(3x-1\right)=x+1\Leftrightarrow y=\dfrac{x+1}{3x-1}\)

\(\left(3x^2+1\right)\left(3+1\right)\ge\left(3x+1\right)^2\Rightarrow\sqrt{3x^2+1}\ge\dfrac{1}{2}\left(3x+1\right)\)

\(\Rightarrow\dfrac{2}{\sqrt{3x^2+1}}\le\dfrac{4}{3x+1}\)

\(\Rightarrow A\le\dfrac{4}{3x+1}+\dfrac{4}{3y+1}=\dfrac{4}{3x+1}+\dfrac{2\left(3x-1\right)}{3x+1}=\dfrac{6x+2}{3x+1}=2\)

\(A_{min}=2\) khi \(x=y=1\)

Bai nay kho qua

cho x+y = 1.Tìm GTLN của Q=x³y+xy³

Q = x³y+xy³

= x.(x²y+y³)

= x.[y.(x²+y²)]              (1)

 x+y = 1

( x + y )²= 1² =1

( x + y )( x + y )=1

( x + y ).x+( x + y ).y=1

x²+y+y²+x = 1  

(x²+y²)+(x+y)=1 maX+Y=1

X²+Y²=0

Thay vao (1 ) ta co

x.[y.(x²+y²)]

=x.(y.0)

=x.0

=0

\(\text{Ta có : }x+y=1\Rightarrow\left\{{}\begin{matrix}1-y=x\\y-1=-x\end{matrix}\right.\left(1\right)\\ \)

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

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

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

Thay \(\left(1\right)\) vào suy ra :

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

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

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

\(A=x-x^2-\left(xy^2-xy^2\right)\)

\(A=x-x^2\)

Mà \(x^2\ge0\)

\(\Rightarrow A=x-x^2\le x\)

Dấu \("="\) xảy ra khi : \(x^2=0\Rightarrow x=0\)

\(\Rightarrow A=x-x^2\le0\)

Vậy \(A_{\left(max\right)}=0\) khi \(x=0\)