\(\frac{12x-15y}{7}=\frac{20z-12x}{9}=\frac{15y-20z}{11}=\frac{12x-15y+20z-12x+15y-20z}{7+9+11}=0\)

\(12x-15y=0\Rightarrow4x=5y\Rightarrow\frac{x}{5}=\frac{y}{4}\)

\(20z-12x=0\Rightarrow5z=3x\Rightarrow\frac{z}{3}=\frac{x}{5}\)

\(15y-20z=0\Rightarrow3y=4z\Rightarrow\frac{y}{4}=\frac{z}{3}\)

\(\Rightarrow\frac{x}{5}=\frac{y}{4}=\frac{z}{3}=\frac{x+y+z}{5+4+3}=\frac{48}{12}=4\)

ta có;x=4x5=20

y=4x4=16

z=4x3=12

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

\(\dfrac{12x-15y+20z-12x+15y-20z}{7+9+11}=0\)

⇒\(\left\{{}\begin{matrix}12x=15y\\20z=12x\\15y=20z\end{matrix}\right.\)

⇔\(12x=15y=20z\)⇒\(\dfrac{x}{5}=\dfrac{y}{4}=\dfrac{z}{3}\)

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

\(\dfrac{x}{5}=\dfrac{y}{4}=\dfrac{z}{3}=\dfrac{x+y+z}{5+4+3}=\dfrac{48}{12}=4\)

⇒\(\left\{{}\begin{matrix}x=5.4=20\\y=4.4=16\\z=3.4=12\end{matrix}\right.\)

(12x-15y)+ (20z-12x)+ (15y-20z)/7+9+11=0

=>12x=15y=>x=5/4y

=>15y=20z=>z=3/4y

x+y+z+48=> y+5/4y+3/4y=48=>y=16

x=5/4*16=20

z=48-20-16=12