x/6-1/y=1/2 
=> quy đồng vế trái đc: (xy-6)/6y=1/2 => xy - 6 = 3y => y*(x-3)=6 
=> có các trường hợp: 
y=1 => x-3=6 
y=2 => x-3=3 
y=3 => x-3=2 
y=6 => x-3=1 
y=-1 => x-3=-6 
y=-2 => x-3=-3 
y=-3 => x-3=-2 
y=-6 => x-3=-1 

=>5=1x5=-1x(-5)

=>x-1=1=1+1   =>x=2                       x-1=5=5+1  =>x=6

    y+2=5=5-2   =>y=3                       y+2=1=1-2  =>y=-1

    x-1=-1=-1+1 =>x=0                       x-1=-5=-5+1 =>x=-4

    y+2=-5=-5-2 =>y=-7                      y+2=-1=-1-2 =>x=-3

   tick cho mình mình tick lại cho

x=1,y=-2

\(\frac{7}{x}=\frac{y}{1}\)

\(\Leftrightarrow x\cdot y=7\)

+) \(\hept{\begin{cases}x=1\\y=7\end{cases}}\)

+) \(\hept{\begin{cases}x=-1\\y=-7\end{cases}}\)

+) \(\hept{\begin{cases}x=7\\y=1\end{cases}}\)

+) \(\hept{\begin{cases}x=-7\\y=-1\end{cases}}\)

Vậy....

A, => x+2=0 hoặc y-3=0

=> x=-2 hoặc y=3

B, => x+1=0 hoặc xy-1=0

=> x=-1 hoặc xy=1

=> x=-1 hoặc x=y=+-1

a)  \(\left(x+2\right).\left(y-3\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x+2=0\\y-3=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=-2\\y=3\end{cases}}\)

       vậy \(\orbr{\begin{cases}x=-2\\y=3\end{cases}}\)

b)  \(\left(x+1\right)\left(xy-1\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x+1=0\\xy-1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=-1\\xy=1\end{cases}}\)

     vậy  \(\orbr{\begin{cases}x=-1\\xy=1\end{cases}}\)

a)Vì \(\left(x+2\right)\left(y-3\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x+2=0\\y-3=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=-2\\y=3\end{cases}}\)

b)Vì \(\left(x+1\right)\left(xy-1\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x+1=0\\xy-1=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=-1\\xy=1\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=-1\\-1.y=1\Leftrightarrow y=-1\end{cases}}\)

x=0 ; y=0

x=1 ; y=2

x=-1 ; y=-1

\(2a=3b\Rightarrow\dfrac{a}{3}=\dfrac{b}{2}\Rightarrow\dfrac{a}{21}=\dfrac{b}{14}\\ 5b=7c\Rightarrow\dfrac{b}{7}=\dfrac{c}{5}\Rightarrow\dfrac{b}{14}=\dfrac{c}{10}\\ \Rightarrow\dfrac{a}{21}=\dfrac{b}{14}=\dfrac{c}{10}\)

Áp dụng t/c dtsbn:

\(\dfrac{a}{21}=\dfrac{b}{14}=\dfrac{c}{10}=\dfrac{3a}{63}=\dfrac{7b}{98}=\dfrac{5c}{50}=\dfrac{3a-7b+5c}{63-98+50}=\dfrac{-30}{15}=-2\\ \Rightarrow\left\{{}\begin{matrix}a=-42\\b=-28\\c=-20\end{matrix}\right.\)

\(x:y:z=3:4:5\Rightarrow\dfrac{x}{3}=\dfrac{y}{4}=\dfrac{z}{5}\)

Đặt \(\dfrac{x}{3}=\dfrac{y}{4}=\dfrac{z}{5}=k\Rightarrow x=3k;y=4k;z=5k\)

\(2x^2+2y^2-3z^2=-100\\ \Rightarrow18k^2+32k^2-75k^2=-100\\ \Rightarrow-25k^2=-100\Rightarrow k^2=4\Rightarrow\left[{}\begin{matrix}k=2\\k=-2\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=6;y=8;z=10\\x=-6;y=-8;z=-10\end{matrix}\right.\)

\(2a=3b\text{⇒}a=\dfrac{3b}{2}\) , \(5b=7c\text{⇒}c=\dfrac{5c}{7}\) 

\(3a-7b+5c\) \(=-30\) 

⇔ \(3.\dfrac{3b}{2}-7b+5.\dfrac{5b}{7}=-30\)  

⇔\(63b-98b+50b=-420\) 

⇔\(b=-28\) ⇒\(\left\{{}\begin{matrix}a=-42\\c=-20\end{matrix}\right.\)

ta có x + y = 2 suy ra x= 2 - y

z + x = -5 suy ra x= -5-z

suy ra x=2 -y = -5 -z=-5-z-2+y= -7 - z + y

thay x=-7 - z + y vào z + x = -5 ta được 

z - 7 -z +y = - 5

-7 + y = -5

y=2

suy ra x= -2 , z=-3