\(a,\dfrac{x}{5}=\dfrac{-18}{10}\\ \Rightarrow x=-\dfrac{18}{10}.5\\ \Rightarrow x=-9\\ b,\dfrac{6}{x-1}=\dfrac{-3}{7}\\ \Rightarrow6.7=-3\left(x-1\right)\\ \Rightarrow42=-3x+3\\ \Rightarrow42+3x-3=0\\ \Rightarrow3x+39=0\\ \Rightarrow3x=-39\\ \Rightarrow x=-13\\ c,\dfrac{y-3}{12}=\dfrac{3}{y-3}\\ \Rightarrow\left(y-3\right)^2=36\\ \Rightarrow\left[{}\begin{matrix}y-2=6\\y-2=-6\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}y=8\\y=-4\end{matrix}\right.\)

\(d,\dfrac{x}{25}=\dfrac{-5}{x^2}\\ \Rightarrow x^3=-125\\ \Rightarrow x^3=\left(-5\right)^3\\ \Rightarrow x=-5\)

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

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

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

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

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)

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

a,\(\dfrac{6}{2x+1}=\dfrac{2}{7}\)

⇒\(\dfrac{6}{2x+1}=\dfrac{6}{21}\)

⇒\(2x+1=21\)

\(2x=21-1\)

\(2x=20\)

⇒\(x=10\)

Giải:

a) \(\dfrac{-1}{5}\le\dfrac{x}{8}\le\dfrac{1}{4}\) 

\(\Rightarrow\dfrac{-8}{40}\le\dfrac{5x}{40}\le\dfrac{10}{40}\) 

\(\Rightarrow5x\in\left\{0;\pm5;10\right\}\) 

\(\Rightarrow x\in\left\{0;\pm1;2\right\}\) 

b) \(\dfrac{4}{x-6}=\dfrac{y}{24}=\dfrac{-12}{18}\) 

\(\Rightarrow\dfrac{4}{x-6}=\dfrac{-12}{18}\) 

\(\Rightarrow-12.\left(x-6\right)=4.18\) 

\(\Rightarrow-12x+72=72\) 

\(\Rightarrow-12x=72-72\) 

\(\Rightarrow-12x=0\) 

\(\Rightarrow x=0:-12\) 

\(\Rightarrow x=0\) 

\(\Rightarrow\dfrac{y}{24}=\dfrac{-12}{18}\) 

\(\Rightarrow y=\dfrac{-12.24}{18}=-16\) 

c) \(\dfrac{x+46}{20}=x.\dfrac{2}{5}\) 

\(\dfrac{x+46}{20}=\dfrac{2x}{5}\) 

\(\Rightarrow5.\left(x+46\right)=2x.20\) 

\(\Rightarrow5x+230=40x\) 

\(\Rightarrow5x-40x=-230\) 

\(\Rightarrow-35x=-230\) 

\(\Rightarrow x=-230:-35\) 

\(\Rightarrow x=\dfrac{46}{7}\) 

Chúc bạn học tốt!

\(\dfrac{1}{x}+\dfrac{y}{3}=\dfrac{5}{6}\Rightarrow\dfrac{6}{6x}+\dfrac{2xy}{6x}=\dfrac{5x}{6x}\Rightarrow6+2xy=5x\)

\(\Rightarrow5x-2xy=6\Rightarrow x\left(5-2y\right)=6\)

Do \(x,y\) là số tự nhiên nên \(x\inƯ^+\left(6\right)\)

TH1: \(x=1\Rightarrow5-2y=6\Rightarrow y=-\dfrac{1}{2}\) (loại)

TH2: \(x=2\Rightarrow5-2y=3\Rightarrow y=1\) (TM)

TH3: \(x=3\Rightarrow5-2y=2\Rightarrow y=\dfrac{3}{2}\) (Loại)

TH4: \(x=6\Rightarrow5-2y=1\Rightarrow y=2\) (TM)

\(\Leftrightarrow6+2xy=5x\left(x\ne0\right)\)

\(\Leftrightarrow5x-2xy=6\Leftrightarrow x\left(5-2y\right)=6\)

\(\Leftrightarrow x=\dfrac{6}{5-2y}\) 

Để x nguyên thì 5-2y phải là ước của 6

\(\Rightarrow5-2y=\left\{-6;-3;-2;-1;1;2;3;6\right\}\)

\(\Rightarrow y=\left\{4;3;2;1\right\}\Rightarrow x=\left\{-2;-6;6;2\right\}\)

\(\Leftrightarrow\dfrac{xy-12}{4y}=\dfrac{5}{8}\)

=>2(xy-12)=5y

=>2xy-24=5y

=>2xy-5y=24

=>y(2x-5)=24

mà x,y là số nguyên

nên \(\left(2x-5;y\right)\in\left\{\left(1;24\right);\left(-1;-24\right);\left(3;8\right);\left(-3;-8\right)\right\}\)

=>\(\left(x,y\right)\in\left\{\left(3;24\right);\left(2;-24\right);\left(4;8\right);\left(1;-8\right)\right\}\)

y = 8

x = 4