tìm x, y, z biết 3x-2y/37 = 5y-3z/15 = 2z-5x/2 và 10x-3y-2z = -4 ( giải bằng cách lập phương trình)

Áp dụng tính chất dãy các tỉ số bằng nhau, ta có:

`(3x – 2y)/37 = (5y – 3z)/15 = (2z – 5x)/2`

`= (5(3x – 2y))/(5. 37) = (2(5y – 3z))/(2. 15) = (3(2z – 5x))/(3. 2)`

`= (15x – 10y)/185 = (10y – 6z)/30 = (6z – 15x)/6`

`= (15x – 10y + 10y – 6z + 6z – 15x)/(185 + 30 + 6)`

`= 0/221`

`= 0`

`<=> (3x – 2y)/37 = (5y – 3z)/15 = (2z – 5x)/2 = 0`
`<=> 3x – 2y = 5y – 3z = 2z – 5x = 0`
`<=>` \(\left\{\begin{matrix}3x – 2y = 0\\5y – 3z = 0\\2z – 5x = 0\end{matrix}\right.\)

`<=>` \(\left\{\begin{matrix}3x = 2y\\5y = 3z\\2z = 5x\end{matrix}\right.\)

`<=>` \(\left\{\begin{matrix}\dfrac{x}{2} = \dfrac{y}{3}\\\dfrac{y}{3} = \dfrac{z}{5} \\\dfrac{z}{5} = \dfrac{x}{2}\end{matrix}\right.\)

`<=> x/2 = y/3 = z/5`
`<=>` \(\left\{\begin{matrix}\dfrac{x}{2} =  \dfrac{y}{3}\\\dfrac{x}{2} =  \dfrac{z}{5}\\\end{matrix}\right.\)

`<=>`  \(\left\{\begin{matrix}y =  \dfrac{3}{2}x\\z =  \dfrac{5}{2}x\end{matrix}\right.\)

Thay  \(\left\{\begin{matrix}y =  \dfrac{3}{2}x\\z =  \dfrac{5}{2}x\end{matrix}\right.\) vào `10x – 3y – 2z = -4`, ta được:

`10x – 3. 3/2x – 2. 5/2x = -4`

`<=> 10x – 9/2x – 5x = -4`

`<=> 1/2x = -4`

`<=> x = -8`

 Với `x = -8 <=>` \(\left\{\begin{matrix}y =  \dfrac{3}{2}. -8 = -12\\z =  \dfrac{5}{2}. -8 = -20\end{matrix}\right.\)

Vậy \(\left\{\begin{matrix}x = -8\\y = -12\\z = -20\end{matrix}\right.\)