Question

Tìm \(x,y,z\), biết: \(\dfrac{x}{4}=\dfrac{y}{5};\dfrac{y}{3}=\dfrac{z}{2}\) và \(x-y+z=21\)

Answer 1

ta có

\(\dfrac{x}{4}=\dfrac{y}{5}\Rightarrow\dfrac{x}{12}=\dfrac{y}{15}\)

\(\dfrac{y}{3}=\dfrac{z}{2}\Rightarrow\dfrac{y}{15}=\dfrac{z}{10}\)

\(\Rightarrow\dfrac{x}{12}=\dfrac{y}{15}=\dfrac{z}{10}\) và \(\text{x − y + z = 21}\)

ADTC dãy tỉ số bằng nhau ta có

\(\dfrac{x}{12}=\dfrac{y}{15}=\dfrac{z}{10}=\dfrac{x-y+z}{12-15+10}=\dfrac{21}{7}=3\)

\(\Rightarrow\left\{{}\begin{matrix}x=12.3=36\\y=15.3=45\\z=10.3=30\end{matrix}\right.\)

Answer 2

`=> x/12 = y/15 = z/10`

Áp dụng t/c dãy tỉ số bằng nhau:

`x/12 = y/15 = z/10 = (x-y+z)/(12-15+10) = 21/7 = 3`

`=> x/12 = 3 => x = 36`

`=> y/15 = 3 => y = 45`

`=> z/10 = 3 => z = 30`

Answer 3

\(\dfrac{x}{4}=\dfrac{y}{5}->x=\dfrac{4}{5}y\) ; \(\dfrac{y}{3}=\dfrac{z}{2}->z=\dfrac{2}{3}y\)

Ta có  \(x-y+z=21\)

\(\Leftrightarrow\) \(\dfrac{4}{5}y-y+\dfrac{2}{3}y=21\)

\(\Leftrightarrow\) \(y\cdot\left(\dfrac{4}{5}-1+\dfrac{2}{3}\right)=21\)

\(\Leftrightarrow\) \(y\cdot\dfrac{7}{15}=21\Rightarrow y=21:\dfrac{7}{15}=45\)

--> \(z=\dfrac{2}{3}y=\dfrac{2}{3}\cdot45=30\)

--> \(x=\dfrac{4}{5}y=\dfrac{4}{5}\cdot45=36\)