\(TH1,\left\{{}\begin{matrix}x-2=2\\2y+3=13\end{matrix}\right.\\ =>\left\{{}\begin{matrix}x=4\\y=5\end{matrix}\right.\\ TH2,\left\{{}\begin{matrix}x-2=13\\2y+3=2\end{matrix}\right.\\ =>\left\{{}\begin{matrix}x=15\\y=-\dfrac{1}{2}\end{matrix}\right.\\ TH3,\left\{{}\begin{matrix}x-2=-2\\2y+3=-13\end{matrix}\right.\\ =>\left\{{}\begin{matrix}x=0\\y=-8\end{matrix}\right.\\ Th4,\left\{{}\begin{matrix}x-2=-13\\2y+3=-2\end{matrix}\right.\\ =>\left\{{}\begin{matrix}x=-11\\y=-\dfrac{5}{2}\end{matrix}\right.\)

x=4

y=5

x=4

y=5

(x-2)(2y+3)=26 

Ta thấy 2y+3 là số lẻ 

x-2                   26             2

2y+3                1              13

x                     28                4

y                      -1                5 

Vậy x=4 ; y= 5

loại y=-1 Vì -1 l;à số nguyên âm 

(x-2)(2y+3)=26=1.26=2.13

\(\Rightarrow\hept{\begin{cases}x-2=1\\2y+3=26\end{cases}\Rightarrow\hept{\begin{cases}x=3\\y=\frac{23}{2}\end{cases}}}\)(loại)

\(\Rightarrow\hept{\begin{cases}x-2=2\\2y+3=13\end{cases}\Rightarrow\hept{\begin{cases}x=4\\y=5\end{cases}}}\)(nhận)

Vậy \(\left(x;y\right)=\left(4;5\right)\)

( x - 2 ) . ( 2y + 3 ) = 26

     x, y \(\in\) N => x - 2 \(\in\) Ư(26)

       Mà x - 2 \(\ge\) 2

=> x - 2 \(\in\) \(\left\{1;2;13;26\right\}\)

Ta có bảng:

Vậy x = 4; y = 5

\(1,=-\left(y^2+12y+36\right)=-y^2-12y-36\)

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

\(3,=-\left(\dfrac{4}{9}+\dfrac{4}{3}x+x^2\right)=-\dfrac{4}{9}-\dfrac{4}{3}x-x^2\)

\(4,=-\left(x^2-3x+\dfrac{9}{4}\right)=-x^2+3x-\dfrac{9}{4}\)

\(5,-\left(2+3y\right)^2=-\left(4+12y+9y^2\right)=-4-12y-9y^2\)

.... mấy ý còn lại bn tự lm nhé, tương tự thhooi

1) \(-\left(y+6\right)^2=-y^2-12y-36\)

2) \(-\left(4-y\right)^2=-y^2+8y-16\)

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

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

5) \(-\left(3y+2\right)^2=-9y^2-12y-4\)

6) \(-\left(2y-3\right)^2=-4y^2+12y-9\)

7) \(-\left(5x+2y\right)^2=-25x^2-20xy-4y^2\)

8) \(-\left(2x-\dfrac{3}{2}\right)^2=-4x^2+6x-\dfrac{9}{4}\)

(x -2) .(2y +3)= 26

=>x -2 ;2y +3\(\in\)U(26)

U(26)={\(\pm1;\pm2;\pm13;\pm26\)}

=>x - 2 ;2y +3\(\in\){\(\pm1;\pm2;\pm13;\pm26\)}

ta có bảng sau:

=>x=-24 thì y=-2

x=28 thì y=-1

x=4 thì y=5

x=0 thì y=-8

(x.2).(2y+3)=26

2x(2y+3)=26

x(2y+3)=13

TH1<=>x=1;2y+3=13

<=>x=1; y=5

TH2 <=> x=13 ;2y+3=1

x=13 ; y= -1

P/s tham khảo nha

x=4;y=5 k min nha