Question

tìm 3 số x,y,z biết: \(\dfrac{x}{2}=\dfrac{y}{3}\) , \(\dfrac{y}{4}=\dfrac{z}{5}\) và x + y -z = 10

Answer 1

Từ \(\dfrac{x}{2}=\dfrac{y}{3}\Rightarrow\dfrac{x}{8}=\dfrac{y}{12}\) và \(\dfrac{y}{4}=\dfrac{z}{5}\Rightarrow\dfrac{y}{12}=\dfrac{z}{15}\)

\(\Rightarrow\dfrac{x}{8}=\dfrac{y}{12}=\dfrac{z}{15}\)

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

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

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{8}=2\Rightarrow x=2\cdot8=16\\\dfrac{y}{12}=2\Rightarrow y=2\cdot12=24\\\dfrac{z}{15}=2\Rightarrow z=2\cdot15=30\end{matrix}\right.\)

Answer 2

Theo bài ta có :

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

\(x+y-z=10\)

\(\Leftrightarrow\dfrac{x}{8}=\dfrac{y}{12}=\dfrac{z}{15}\)

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

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

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x}{8}=2\Leftrightarrow x=16\\\dfrac{y}{12}=2\Leftrightarrow y=24\\\dfrac{z}{15}=2\Leftrightarrow z=30\end{matrix}\right.\)

Vậy ....

Answer 3

\(\dfrac{x}{2}=\dfrac{y}{3}\Rightarrow\dfrac{x}{8}=\dfrac{y}{12}\)và \(\dfrac{y}{4}=\dfrac{z}{5}\Rightarrow\dfrac{y}{12}=\dfrac{z}{15}\)
=> \(\dfrac{x}{8}=\dfrac{y}{12}=\dfrac{z}{15}\)
Ta có :
\(\dfrac{x}{8}=\dfrac{y}{12}=\dfrac{z}{15}=\dfrac{x+y-z}{8+12-15}=\dfrac{10}{5}=2\)
=> \(\left\{{}\begin{matrix}\dfrac{x}{8}=2\Rightarrow x=2.8=16\\\dfrac{y}{12}=2\Rightarrow x=2.12=24\\\dfrac{z}{15}\Rightarrow2.15=30\end{matrix}\right.\)
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