Question

1, Tìm x,y,z, biết

a, \(\dfrac{3x-2y}{4}\) = \(\dfrac{2z-4x}{3}\) = \(\dfrac{4y-3z}{2}\) và x + y + z = 18

Answer 1

Giải

Ta có:

\(\dfrac{3x-2y}{4}=\dfrac{2z-4x}{3}=\dfrac{4y-3z}{2}\)\(=\dfrac{12x-8y+6z-12+8y-6z}{16+9+4}\)\(=\)\(\dfrac{0}{29}=0\)

\(\Rightarrow\left\{{}\begin{matrix}3x-2y=0\Rightarrow3x=2y\Rightarrow\dfrac{x}{2}=\dfrac{y}{3}\\2z-4x=0\Rightarrow2z=4x\Rightarrow\dfrac{x}{2}=\dfrac{z}{4}\end{matrix}\right.\)

\(\Rightarrow\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}\)

Áp dụng tính chất dãy tỉ số bằng nhau:

\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}=\dfrac{x+y+z}{2+3+4}=\dfrac{18}{9}=2\)

\(\Rightarrow\left\{{}\begin{matrix}x=2.2=4\\y=2.3=6\\z=2.4=8\end{matrix}\right.\)

Vậy \(x=4;y=6;z=8\)

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