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

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

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

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

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

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

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

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

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

⇒\(\dfrac{6}{2x+1}=\dfrac{6}{21}\)

⇒\(2x+1=21\)

\(2x=21-1\)

\(2x=20\)

⇒\(x=10\)

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:

ĐKXĐ: x<>0

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

=>\(\dfrac{6+xy}{3x}=\dfrac{1}{6}\)

=>\(6\left(6+xy\right)=3x\)

=>\(x=2\left(6+xy\right)=12+2xy\)

=>\(x\left(1-2y\right)=12\)

mà x,y là các số nguyên

nên \(\left(x;1-2y\right)\in\left\{\left(12;1\right);\left(-12;-1\right);\left(4;3\right);\left(-4;-3\right)\right\}\)

=>\(\left(x,y\right)\in\left\{\left(12;0\right);\left(-12;1\right);\left(4;-1\right);\left(-4;2\right)\right\}\)

c: ĐKXĐ: y<>-1

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

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

=>\(\dfrac{2\left(xy+x+3\right)}{6\left(y+1\right)}=\dfrac{y+1}{6\left(y+1\right)}\)

=>\(2xy+2x+6=y+1\)

=>\(2x\left(y+1\right)-\left(y+1\right)=-6\)

=>\(\left(2x-1\right)\left(y+1\right)=-6\)

mà x,y là các số nguyên

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

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