Question

Giúp với cần gấp x phần y = 9 phần 7; y phần z = 7 phần 3 và x - y + z=-15

Answer 1

\(TC:\dfrac{x}{y}=\dfrac{9}{7}\Rightarrow\dfrac{x}{9}=\dfrac{y}{7}\)

\(\dfrac{y}{z}=\dfrac{7}{3}\Rightarrow\dfrac{y}{7}=\dfrac{z}{3}\)

\(KĐ:\) \(\dfrac{x}{9}=\dfrac{y}{7}=\dfrac{z}{3}=\dfrac{x-y+z}{9-7+3}=\dfrac{-15}{5}=-3\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=-3\cdot9=-27\\y=-3\cdot7=-21\\z=-3\cdot3=-9\end{matrix}\right.\)

Answer 2

 

\(\dfrac{x}{y}=\dfrac{9}{7}\Rightarrow\dfrac{x}{9}=\dfrac{y}{7}\)

\(\dfrac{y}{z}=\dfrac{7}{3}\Rightarrow\dfrac{y}{7}=\dfrac{z}{3}\)

\(\Rightarrow\dfrac{x}{9}=\dfrac{y}{7}=\dfrac{z}{3}\)

AD TC DTSBN ta có 

\(\dfrac{x}{9}=\dfrac{y}{7}=\dfrac{z}{3}=\dfrac{x-y+z}{9-7+3}=\dfrac{-15}{5}=-3\)

\(\Rightarrow\dfrac{x}{9}=-3\Rightarrow x=-27\)

\(\dfrac{y}{7}=-3\Rightarrow y=-21\)

\(\dfrac{z}{3}=-3\Rightarrow z=-9\)

Answer 3

Ta có: \(\dfrac{x}{y}=\dfrac{9}{7}\)

\(\Leftrightarrow\dfrac{x}{9}=\dfrac{y}{7}\)(1)

Ta có: \(\dfrac{y}{z}=\dfrac{7}{3}\)

\(\Leftrightarrow\dfrac{y}{7}=\dfrac{z}{3}\)(2)

Từ (1) và (2) suy ra \(\dfrac{x}{9}=\dfrac{y}{7}=\dfrac{z}{3}\)

mà x-y+z=-15

nên Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

\(\dfrac{x}{9}=\dfrac{y}{7}=\dfrac{z}{3}=\dfrac{x-y+z}{9-7+3}=\dfrac{-15}{5}=-3\)

Do đó:

\(\left\{{}\begin{matrix}\dfrac{x}{9}=-3\\\dfrac{y}{7}=-3\\\dfrac{z}{3}=-3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-27\\y=-21\\z=-9\end{matrix}\right.\)

Vậy: (x,y,z)=(-27;-21;-9)