\(\Rightarrow x=\dfrac{31}{5}\)

Vậy \(x=\dfrac{31}{5}\)

\(\dfrac{8}{7}-\dfrac{1}{7}:\left(\dfrac{x}{3}-2\right)=-1\\ \Rightarrow\dfrac{1}{7}:\left(\dfrac{x}{3}-2\right)=\dfrac{8}{7}+1\\ \Rightarrow\dfrac{1}{7}:\left(\dfrac{x}{3}-2\right)=\dfrac{15}{7}\\ \Rightarrow\dfrac{x}{3}-2=\dfrac{1}{7}:\dfrac{15}{7}\\ \Rightarrow\dfrac{x}{3}-2=\dfrac{1}{15}\\ \Rightarrow\dfrac{x}{3}=\dfrac{1}{15}+2\\ \Rightarrow\dfrac{x}{3}=\dfrac{31}{15}\\ \Rightarrow15x=31.3\\ \Rightarrow15x=93\\ \Rightarrow x=\dfrac{93}{15}\)

`{((x-3)(y+4)=xy-4),((x-1)(y+2)=xy+6):}`

`1+2-3+4-5+6-7+8-9+10`

`= (1+10) + (2-3) + (4-5) + (6-7) + (8-9)`

`= 11 + (-1) + (-1) +(-1) +(-1)`

`= 11 + (-1) xx 4`

`= 11 -4`

`=7`