Question

[Nguyễn Lê Thảo Nguyên]

Tìm x y z biết \(\dfrac{x}{y}=\dfrac{3}{5}\) , 7y=6z và 4x + 8y - 9z = -3

Answer 1

[Trần Minh Hoàng]

Ta có:

\(\dfrac{x}{y}=\dfrac{3}{5}\Rightarrow\dfrac{x}{18}=\dfrac{y}{30}\)

\(7y=6z\Rightarrow\dfrac{y}{30}=\dfrac{z}{35}\)

\(\Rightarrow\dfrac{x}{18}=\dfrac{y}{30}=\dfrac{z}{35}\)

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

\(\Rightarrow\dfrac{x}{18}=\dfrac{y}{30}=\dfrac{z}{35}=\dfrac{4x}{72}=\dfrac{8y}{240}=\dfrac{9z}{315}=\dfrac{4x+8y-9z}{72+240-315}=\dfrac{-3}{-3}=1\)

\(\Rightarrow\left\{{}\begin{matrix}x=18\\y=30\\z=35\end{matrix}\right.\)

Vậy...

Answer 2

[Nguyễn Nam]

Ta có: \(\dfrac{x}{y}=\dfrac{3}{5}\Rightarrow\dfrac{x}{3}=\dfrac{y}{5}\) (1)

\(7y=6z\Rightarrow\dfrac{y}{6}=\dfrac{z}{7}\) (2)

Từ (1) và (2) suy ra: \(\dfrac{x}{3}=\dfrac{y}{5};\dfrac{y}{6}=\dfrac{z}{7}\Leftrightarrow\dfrac{x}{18}=\dfrac{y}{30};\dfrac{y}{30}=\dfrac{z}{35}\Rightarrow\dfrac{x}{18}=\dfrac{y}{30}=\dfrac{z}{35}\)

Có \(\dfrac{x}{18}=\dfrac{y}{30}=\dfrac{z}{35}\)và \(4x+8y-9z=-3\)

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

\(\dfrac{x}{18}=\dfrac{y}{30}=\dfrac{z}{35}\Rightarrow\dfrac{4x}{72}=\dfrac{8y}{240}=\dfrac{9z}{315}=\dfrac{4x+8y-9z}{72+240-315}=\dfrac{-3}{-3}=1\)

\(\dfrac{4x}{72}=1\Rightarrow4x=72\Rightarrow x=\dfrac{72}{4}=18\)

\(\dfrac{8y}{240}=1\Rightarrow8y=240\Rightarrow y=\dfrac{240}{8}=30\)

\(\dfrac{9z}{315}=1\Rightarrow9z=315\Rightarrow z=\dfrac{315}{9}=35\)

Vậy x=18 ; y=30 ; z=35