=> 1/x = 5/6 - y/3

1/x = 5-2y/6

=> x(5-2y) = 1.6 = 6

Do x ∈ N => x >= 0

Mà 6>0 => 5-2y > 0

Vì y ∈ N => 5-2y ∈ N*

Ta có bảng:

Do x,y ∈ N => (x,y) = (6,2) (thử lại thỏa mãn)

Vậy x=6; y = 2

=>(12-xy)/3x=5/6

=>6(12-xy)=15x

=>(12-xy)=5/2x

=>24-2xy=5x

=>5x+2xy=24

=>x(2y+5)=24

=>(x;2y+5) thuộc {(1;24); (2;12); (3;8); (4;6); (6;4); (8;3); (12;2); (24;1)}

mà x,y là các số tự nhiên

nên \(\left(x,y\right)\in\varnothing\)

\(\dfrac{x+1}{2}=\dfrac{y-5}{3}=\dfrac{z-4}{4}=\dfrac{x+1+y-5-z+4}{2+3-4}\)

\(=\dfrac{7}{1}=7\)

\(\Rightarrow\left\{{}\begin{matrix}x=7.2-1=13\\y=7.3+5=26\\z=7.4+4=32\end{matrix}\right.\)

Áp dụng t/c dtsbn:

\(\dfrac{x+1}{2}=\dfrac{y-5}{3}=\dfrac{z-4}{4}=\dfrac{x+1+y-5-z+4}{2+3-4}=\dfrac{7+1+4-5}{1}=7\\ \Rightarrow\left\{{}\begin{matrix}x+1=14\\y-5=21\\z-4=28\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=13\\y=26\\z=32\end{matrix}\right.\)

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

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

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

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

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

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

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

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

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

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

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

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