1) Tìm cặp số nguyên ($x; y$) sao cho: $x+y=xy+3$ 2) Cho $x+y=3$ Tìm GTLN của A= $xy$

Đáp án+Giải thích các bước giải:

 `1)`

`x+y=xy+3`

`<=>x+y-xy-3=0`

`<=>x+y-xy-1-2=0`

`<=>(x-1)+(y-xy)=2`

`<=>1.(x-1)+y.(1-x)=2`

`<=>1.(x-1)-y.(x-1)=2`

`<=>(x-1).(1-y)=2`

`=>` \begin{array}{|c|c|c|}\hline \text{x - 1}&\text{1}&-1&\text{2}&\text{-2}\\\hline \text{1 - y}&\text{2}&\text{-2}&\text{1}&\text{-1}\\\hline \text{x}&\text{2}&\text{0}&\text{3}&\text{-1}\\\hline \text{y}&\text{-1}&\text{3}&\text{0}&\text{2}\\\hline \end{array}

Vậy `(x;y)={(2;-1);(0;3);(3;0);(-1;2)}`

`2)`

`x+y=3`

`=>y=3-x`

`A=xy`

`=>A=x.(3-x)`

`=>A=3x-x^{2}`

`=>A=-x^{2}+3x`

`=>A=-(x^{2}-3x)`

`=>A=-[x^{2}-2.x. 3/2+(3/2)^{2}-9/4]`

`=>A=-[(x-3/2)^{2}-9/4]`

`=>A=-(x-3/2)^{2}+9/4`

Ta có:

`(x-3/2)^{2}\ge0AAx`

`=>-(x-3/2)^{2}\le0AAx`

`=>-(x-3/2)^{2}+9/4\le 9/4<0AAx`

`=>` `GTL N_{A}=9/4`

`-)` Dấu $\text{"="}$ xảy ra khi và chỉ khi:

`x-3/2=0<=>x=3/2`

Vậy `GTLN_{A}=9/4` khi `x=3/2`