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

=>3(x+y)=xy

=>3x+3y=xy

=>3x=xy-3y

=>3x=y(x-3)

=>y=\(\dfrac{3x}{x-3}\)

* Vì y nguyên nên 3x ⋮ x-3 

=>3(x-3)+9 ⋮x-3

=>9 ⋮ x-3

=>x-3∈Ư(9)

=>x-3∈{1;-1;3;-3;9;-9}

=>x∈{4;2;6;0;12;-6} mà x nguyên dương và x khác 0 nên x∈{4;2;6;12}

=>y∈{12;-6;6;4} mà y nguyên dương nên y∈{12;6;4}

=>x∈{4;6;12}

- Vậy x=4 thì y=12 ; x=6 thì y=6 ; x=12 thì y=4.

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

\(\dfrac{x}{6}-\dfrac{5.2}{2y.2+1.2}=\dfrac{4}{6}\)(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

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

Đẳng thức xảy ra khi x = y = z = 1.

a, \(\dfrac{x}{2}=-\dfrac{5}{y}\Rightarrow xy=-10\Rightarrow x;y\inƯ\left(-10\right)=\left\{\pm1;\pm2;\pm5;\pm10\right\}\)

c, \(\dfrac{3}{x-1}=y+1\Rightarrow\left(y+1\right)\left(x-1\right)=3\Rightarrow x-1;y+1\inƯ\left(3\right)=\left\{\pm1;\pm3\right\}\)

b: =>xy=12

\(\Leftrightarrow\left(x,y\right)\in\left\{\left(12;1\right);\left(6;2\right);\left(4;3\right)\right\}\)

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

Bài 1: 

b) ĐKXĐ: \(x\ne3\)

Ta có: \(\dfrac{3-x}{20}=\dfrac{-5}{x-3}\)

\(\Leftrightarrow\dfrac{x-3}{-20}=\dfrac{-5}{x-3}\)

\(\Leftrightarrow\left(x-3\right)^2=100\)

\(\Leftrightarrow\left[{}\begin{matrix}x-3=10\\x-3=-10\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=13\left(nhận\right)\\x=-7\left(nhận\right)\end{matrix}\right.\)

Vậy: \(x\in\left\{13;-7\right\}\)

b, Ta có : \(\dfrac{x}{3}=\dfrac{y}{4};\dfrac{y}{5}=\dfrac{z}{6}\Rightarrow\dfrac{x}{15}=\dfrac{y}{20}=\dfrac{z}{24}\)

Đặt \(x=15k;y=20k;z=24k\)

Thay vào A ta được : \(A=\dfrac{30k+60k+96k}{45k+80k+120k}=\dfrac{186k}{245k}=\dfrac{186}{245}\)

lk

a, \(\dfrac{x}{7}-\dfrac{1}{2}=\dfrac{y}{y+1}\Leftrightarrow\dfrac{2x-7}{14}=\dfrac{y}{y+1}\Rightarrow\left(2x-7\right)\left(y+1\right)=14y\)

\(\Leftrightarrow2xy+2x-7y-7=14y\Leftrightarrow2xy+2x-21y-7=0\)

\(\Leftrightarrow2x\left(y+1\right)-21\left(y+1\right)+14=0\Leftrightarrow\left(2x-21\right)\left(y+1\right)=-14\)

\(\Rightarrow2x-21;y+1\inƯ\left(-14\right)=\left\{\pm1;\pm2;\pm7;\pm14\right\}\)

\(\dfrac{1}{x}-\dfrac{1}{y}=\dfrac{1}{x-y}\left(ĐK:x>0;y>0\right)\)

\(\Rightarrow\dfrac{y-x}{xy}=\dfrac{1}{x-y}\)

\(\Rightarrow\left(y-x\right)\left(x-y\right)=xy\)

\(\Rightarrow-\left(x-y\right)^2=xy\) \(^{\left(1\right)}\)

Vì x, y nguyên dương khác nhau và khác 0 ⇒ \(xy>0 \) \(^{\left(2\right)}\)

Ta thấy: \(\left(x-y\right)^2>0\forall x;y\in Z;x\ne y\)

\(\Rightarrow-\left(x-y\right)^2< 0\forall x;y\in Z;x\ne y\)  \(^{\left(3\right)}\)

Từ \(\left(1\right);\left(2\right)\) và \(\left(3\right)\) \(\Rightarrow\) Không tìm được hai số x, y nguyên dương khác nhau thoả mãn yêu cầu đề bài.

#\(Urushi\)