Question

[gtrutykyu]

Tìm x, y,z biết \(\dfrac{x}{3}=\dfrac{y-5}{7}=\dfrac{z+2}{3}\&x+2y=5z\)

b/ \(\dfrac{2x-y}{5}=\dfrac{3y-2z}{15}\&x+z=2y\)

Answer 1

[Mashiro Shiina]

\(\dfrac{x}{3}=\dfrac{y-5}{7}=\dfrac{z+2}{3}\)

\(\Leftrightarrow\dfrac{x}{3}=\dfrac{2y-10}{14}=\dfrac{5z+10}{15}\)

\(x+2y=5z\Leftrightarrow x+2y-5z=0\)

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

\(\dfrac{x}{3}=\dfrac{2y-10}{14}=\dfrac{5z+10}{15}=\dfrac{x+2y-10-5z-10}{3+14-15}\)

\(=\dfrac{-20}{2}=-10\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=-30\\y=-65\\z=-32\end{matrix}\right.\)

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