\(\dfrac{x}{3}\) + \(\dfrac{1}{2}\) = \(\dfrac{1}{y+3}\)  Đk (\(y\ne-3\))⇒ \(\dfrac{2x+3}{6}\) = \(\dfrac{1}{y+3}\) ⇒ (2\(x\)+3)(y+3) = 6

Ư(6) = { -6; -3; -2; -1; 1; 2; 3; 6}

Lập bảng ta có:

Từ bảng trên ta có các cặp \(x\), y nguyên thỏa mãn đề bài là:

(\(x\), y) = ( -3; -5); ( -2; -9); ( -1; 3); (0; -1); 

(2,1),(1,5),

Lời giải:
$\frac{2}{x}+\frac{y}{3}=\frac{1}{6}$

$\frac{6+xy}{3x}=\frac{1}{6}$

$\frac{2(6+xy)}{6x}=\frac{x}{6x}$

$\Rightarrow 2(6+xy)=x$

$\Rightarrow 12+2xy-x=0$

$12=x-2xy$

$12=x(1-2y)$

$\Rightarrow 1-2y$ là ước của $12$

Mà $1-2y$ lẻ nên $1-2y$ là ước lẻ của $12$

$\Rightarrow 1-2y\in\left\{\pm 1; \pm 3\right\}$

$\Rightarrow y\in\left\{0; 1; 2; -1\right\}$

$\Rightarrow x\in\left\{12; -12; -4; 4\right\}$ (tương ứng)

\(\dfrac{x}{6}-\dfrac{5}{2y+1}=\dfrac{2}{3}\)

\(\dfrac{x}{6}-\dfrac{5.2}{2y.2+1.2}=\dfrac{4}{6}\)(vì 2y + 1 là số lẻ)

\(\dfrac{x}{6}-\dfrac{10}{4y+2}=\dfrac{4}{6}\)

Để \(\dfrac{x}{6}-\dfrac{10}{4y+2}=\dfrac{4}{6}\)thì y = 1 để cùng mẫu số

Khi đó ta có\(\dfrac{x}{6}-\dfrac{10}{4y+2}=\dfrac{4}{6}\) = \(\dfrac{x}{6}-\dfrac{10}{4+2}=\dfrac{4}{6}\) = \(\dfrac{x}{6}-\dfrac{10}{6}=\dfrac{4}{6}\)

Vì 4+10 = 14 => x = 14

Vậy y = 1; x = 14

Ta có: \(\frac{x}{6}-\frac{5}{2y+1}=\frac23\)

=>\(\frac{x}{6}-\frac23=\frac{5}{2y+1}\)

=>\(\frac{x-4}{6}=\frac{5}{2y+1}\)

=>(x-4)(2y+1)=30

mà 2y+1 lẻ và 2y+1>1(do x,y là các số nguyên dương)

nên (x-4;2y+1)∈{(10;3);(6;5);(2;15)}

=>(x;y)∈{(14;1);(10;2);(6;7)}

Ta có: \(\frac{x-1}{9}+\frac13=\frac{1}{y+2}\)

=>\(\frac{x-1+3}{9}=\frac{1}{y+2}\)

=>(x+2)(y+2)=9

=>(x+2;y+2)∈{(1;9);(9;1);(-1;-9);(-9;-1);(3;3);(-3;-3)}

=>(x;y)∈{(-1;7);(7;-1);(-3;-11);(-11;-3);(1;1);(-5;-5)}

\(\dfrac{x}{3}-\dfrac{2}{y}=\dfrac{1}{2}\\ \Rightarrow\dfrac{2}{y}=\dfrac{x}{3}-\dfrac{1}{2}\\\Rightarrow \dfrac{2}{y}=\dfrac{2x-3}{6}\\ \Rightarrow y\left(2x-3\right)=2\cdot6\\ \Rightarrow y\left(2x-3\right)=12\)

mà `y in ZZ;x in ZZ`

`=>y in ZZ;2x-3 in ZZ`

`=>y;2x-3` thuộc ước nguyên của `12`

`=>y;2x-3 in {+-1;+-2;+-3;+-4;+-6;+-12}`

Ta có bảng sau :

Vì `x;y in ZZ`

nên `(x;y)=(1;-1);(0;-3);(2;1);(3;3)`

=>\(\dfrac{xy+x-3}{3\left(y+1\right)}=\dfrac{1}{6}\)

=>2(xy+x-3)=y+1

=>2xy+2x-6-y-1=0

=>2x(y+1)-y-1=6

=>(y+1)(2x-1)=6

=>\(\left(2x-1;y+1\right)\in\left\{\left(1;6\right);\left(6;1\right);\left(-1;-6\right);\left(-6;-1\right);\left(2;3\right);\left(3;2\right);\left(-2;-3\right);\left(-3;-2\right)\right\}\)

=>\(\left(x,y\right)\in\left\{\left(1;5\right);\left(\dfrac{7}{2};0\right);\left(-1;-7\right);\left(-\dfrac{5}{2};-1\right);\left(\dfrac{3}{2};2\right);\left(2;1\right);\left(-\dfrac{1}{2};-4\right);\left(-1;-3\right)\right\}\)

\(\dfrac{x}{x^2+yz}+\dfrac{y}{y^2+zx}+\dfrac{z}{z^2+xy}\le\dfrac{x}{2\sqrt{x^2yz}}+\dfrac{y}{2\sqrt{y^2zx}}+\dfrac{z}{2\sqrt{z^2xy}}=\dfrac{1}{2}\left(\dfrac{1}{\sqrt{yz}}+\dfrac{1}{\sqrt{zx}}+\dfrac{1}{\sqrt{xy}}\right)\le\dfrac{1}{2}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)=\dfrac{3}{2}\).

Đẳng thức xảy ra khi x = y = z = 1.

a: \(x^2+y^2(x-y+1)-\left(x-1\right)\cdot y=22\)

=>\(x^2 + y^2(x - y + 1) - xy + y = 22\)

=>\(x^2-xy+y^2(x-y+1)+y=22\)

=>\(x(x-y+1)-x+y^2(x-y+1)+y=22\)

\(\iff(x-y+1)(x+y^2)-(x-y)=22\)

\(\iff(x-y+1)(x+y^2)-(x-y+1)+1=22\)

=>\((x-y+1)(x+y^2-1)=21\)

TH1: \(\begin{cases}x-y+1=1\\ x+y^2-1=21\end{cases}\)

=>\(\begin{cases}x=y\\ y+y^2-1=21\end{cases}\iff y^2+y-22=0\)

=>\(y^2+y+\frac14-\frac{89}{4}=0\)

=>\(\left(y+\frac12\right)^2=\frac{89}{4}\)

=>\(\left[\begin{array}{l}y+\frac12=\frac{\sqrt{89}}{2}\\ y+\frac12=-\frac{\sqrt{89}}{2}\end{array}\right.\Rightarrow\left[\begin{array}{l}y=\frac{\sqrt{89}-1}{2}\left(loại\right)\\ y=\frac{-\sqrt{89}-1}{2}\left(loại\right)\end{array}\right.\)

=>Loại

TH2: \(\begin{cases}x-y+1=21\\ x+y^2-1=1\end{cases}\iff\begin{cases}x=y+20\\ (y+20)+y^2-1=1\end{cases}\iff y^2+y+18=0\)

=>\(y^2+y+\frac14+\frac{71}{4}=0\)

=>\(\left(y+\frac12\right)^2+\frac{71}{4}=0\) (vô lý)

=>Loại

TH3: \(\begin{cases}x-y+1=3\\ x+y^2-1=7\end{cases}\iff\begin{cases}x=y+2\\ (y+2)+y^2-1=7\end{cases}\iff y^2+y-6=0\)

=>(y+3)(y-2)=0

=>y=-3 hoặc y=2

Nếu y=-3 thì x-y+1=3

=>x-y=2

=>x-(-3)=2

=>x+3=2

=>x=-1

Nếu y=2 thì x-y+1=3

=>x-2+1=3

=>x-1=3

=>x=4

TH4: \(\begin{cases}x-y+1=7\\ x+y^2-1=3\end{cases}\iff\begin{cases}x=y+6\\ (y+6)+y^2-1=3\end{cases}\iff y^2+y+2=0\)

=>y∈∅

=>Loại