Question

Tìm x, y, z biết: (x+y):(5-z):(y+z):(9+y) = 3:1:2:5

Answer 1

Ta có: \(\dfrac{x+y}{3}=\dfrac{5-z}{1}=\dfrac{y+z}{2}=\dfrac{9+y}{5}=k\)

\(\dfrac{\left(x+y\right)+\left(5-z\right)+\left(y-z\right)+\left(9-y\right)}{3+1+2+5}=\dfrac{x+y-4}{1}\)

⇒\(\left\{{}\begin{matrix}x+y-4\\x+y=3k\end{matrix}\right.\Rightarrow k+4=x+y\)

\(\Rightarrow4+k=3k\Rightarrow4=2k\Rightarrow k=2\)

\(\Rightarrow5-z=k\Rightarrow z=5-k=5-2=3\)

\(\Rightarrow9+y=5k\Rightarrow9+y=10\Rightarrow y=10-9=1\)

\(\Rightarrow x+y=3k\Rightarrow x+y=6\Rightarrow x=6-y=6-1=5\)

Vậy \(\left\{{}\begin{matrix}x=5\\y=1\\z=3\end{matrix}\right.\)

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Answer 2