`-3x=2y `

`=> x/2 = -y/3 `

AD t/c của dãy tỉ số bằng nhau ta có

`x/2 =-y/3 = (x-y)/(2+3) = 6/5`

`=>{(x=2*6/5 = 12/5),(y=-3*6/5 =-18/5):}`

a) `6/x =-3/2`

`=>x =6 :(-3/2) = 6*(-2/3)=-4`

`b)`\(-3x=2y\Rightarrow\dfrac{x}{2}=\dfrac{y}{-3}\)

Áp dụng t/c của DTSBN , ta đc :

\(\dfrac{x}{2}=\dfrac{y}{-3}=\dfrac{x-y}{2+3}=\dfrac{6}{5}\\ \Rightarrow\left\{{}\begin{matrix}\dfrac{x}{2}=\dfrac{6}{5}\\\dfrac{y}{-3}=\dfrac{6}{5}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\dfrac{12}{5}\\y=-\dfrac{18}{5}\end{matrix}\right. \)

`a)`

`6/x=-3/2`

`x=6:(-3/2)`

`x=6*(-2/3)`

`x=-4`

a, \(\dfrac{6}{x}=-\dfrac{3}{2}\)

\(\Rightarrow3x=6.\left(-2\right)\)

\(\Rightarrow3x=-12\)

\(\Rightarrow x=-\dfrac{12}{3}\)

=>x=-4

b, \(+\left(-3\right)x=2y\Rightarrow\dfrac{x}{2}=\dfrac{y}{-3}\)

+x-y=6

+Áp dụng tính chất dãy tỉ số bằng nhau có:

 \(\dfrac{x}{2}=\dfrac{y}{-3}=\dfrac{x-y}{2-\left(-3\right)}=\dfrac{6}{5}\)

Suy ra \(\dfrac{x}{2}=\dfrac{6}{5}\Rightarrow x=\dfrac{6}{5}.2\Rightarrow x=\dfrac{12}{5}\)

           \(\dfrac{y}{-3}=\dfrac{6}{5}\Rightarrow y=\dfrac{6}{5}.\left(-3\right)=-\dfrac{18}{5}\)

Vậy \(x=\dfrac{12}{5};y=-\dfrac{18}{5}\)

Ta thấy $x^2+y^2+z^2\geq 0$ với mọi $x,y,z$

Do đó $x^2+y^2+z^2=-14$ là vô lý

PT vô nghiệm.

     \(x^2+y^2+z^2=4x-2y+6=-14\)

⇔ \(x^2-4x+4+y^2+2y+1+z^2-6z+9=0\)

⇔ \(\left(x-2\right)^2+\left(y+1\right)^2+\left(z-3\right)^2=0\)

⇔ \(\left\{{}\begin{matrix}\left(x-2\right)^2\\\left(y+1\right)^2\\\left(z-3\right)^2\end{matrix}\right.\) ⇔ \(\left\{{}\begin{matrix}x=2\\y=-1\\z=3\end{matrix}\right.\)

\(\dfrac{19}{21}< \dfrac{x}{y}< \dfrac{7}{6}\Rightarrow\dfrac{38}{42}< \dfrac{x}{y}< \dfrac{49}{42}\Rightarrow\dfrac{x}{y}\in\left\{\dfrac{13}{14};\dfrac{20}{21};\dfrac{41}{42}\right\}\)

Xét \(\dfrac{x}{y}=\dfrac{13}{14}\Rightarrow\dfrac{x}{13}=\dfrac{y}{14}=\dfrac{3x-2y}{39-28}=\dfrac{5}{11}\Rightarrow\left\{{}\begin{matrix}x=\dfrac{65}{11}\\y=\dfrac{70}{11}\end{matrix}\right.\)

Xét \(\dfrac{x}{y}=\dfrac{20}{21}\Rightarrow\dfrac{x}{20}=\dfrac{y}{21}=\dfrac{3x-2y}{60-42}=\dfrac{5}{18}\Rightarrow\left\{{}\begin{matrix}x=\dfrac{50}{9}\\y=\dfrac{35}{6}\end{matrix}\right.\)

Xét \(\dfrac{x}{y}=\dfrac{41}{42}\Rightarrow\dfrac{x}{41}=\dfrac{y}{42}=\dfrac{3x-2y}{123-84}=\dfrac{5}{39}\Rightarrow\left\{{}\begin{matrix}x=\dfrac{205}{39}\\y=\dfrac{70}{13}\end{matrix}\right.\)