Question

Tìm x và y biết :

\(\dfrac{x}{2}\)=\(\dfrac{y}{3}\);\(\dfrac{y}{4}\)=\(\dfrac{z}{5}\) và x+y-z=10

Answer 1

\(\dfrac{x}{2}=\dfrac{y}{3}\Rightarrow\dfrac{x}{8}=\dfrac{y}{12}\)

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

\(\Rightarrow\dfrac{x}{8}=\dfrac{y}{12}=\dfrac{z}{15}=\dfrac{x+y-z}{8+12-15}=\dfrac{10}{5}=2\)

\(\Rightarrow\left\{{}\begin{matrix}x=8.2=16\\y=12.2=24\\z=15.2=30\end{matrix}\right.\)

Answer 2

Ta có \(\dfrac{x}{2}=\dfrac{y}{3}\Rightarrow\dfrac{1}{4}\cdot\dfrac{x}{2}=\dfrac{1}{4}\cdot\dfrac{y}{3}\Rightarrow\dfrac{x}{8}=\dfrac{y}{12}\left(1\right)\)

\(\dfrac{y}{4}=\dfrac{z}{5}\Rightarrow\dfrac{1}{3}\cdot\dfrac{y}{4}=\dfrac{1}{3}\cdot\dfrac{z}{5}\Rightarrow\dfrac{y}{12}=\dfrac{z}{15}\left(2\right)\)

Từ ( 1 ) và ( 2 ) ta có

\(\dfrac{x}{8}=\dfrac{y}{12}=\dfrac{z}{15}\) và x+y-z=10

Áp dụng t/c dãy tỉ số bằng nhau ta có:

\(\dfrac{x}{8}=\dfrac{y}{12}=\dfrac{z}{15}=\dfrac{x+y-z}{8+12-15}=\dfrac{10}{5}=2\)

\(\Rightarrow\dfrac{x}{8}=2\Rightarrow x=2\cdot8=16\)

\(\dfrac{y}{12}=2\Rightarrow y=2\cdot12=24\)

\(\dfrac{z}{15}=2\Rightarrow z=2\cdot15=30\)

vậy x = 16; y = 24; z = 30

Chúc bn học tốt

Answer 3

Ta có: \(\dfrac{x}{2}=\dfrac{y}{3}\) => \(\dfrac{x}{8}=\dfrac{y}{12}\) => \(\dfrac{x}{8}=\dfrac{y}{12}=\dfrac{z}{15}\)

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

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

\(\dfrac{x}{8}=\dfrac{y}{12}=\dfrac{z}{15}=\dfrac{x+y-z}{8+12-15}=\dfrac{10}{5}=2\)

=> x = 2.8 => x = 16

y = 2.12 y = 24

z = 2.15 z = 30