Question

Tìm x,y,z

\(\dfrac{x}{5}\) =\(\dfrac{y}{6}\) ;\(\dfrac{y}{8}\) =\(\dfrac{z}{7}\) và x+y-z=6 chín

Answer 1

Ta có:

\(\dfrac{x}{5}=\dfrac{y}{6}\Rightarrow\dfrac{x}{20}=\dfrac{y}{24}\)

\(\dfrac{y}{8}=\dfrac{z}{7}\Rightarrow\dfrac{y}{24}=\dfrac{z}{21}\)

\(\Rightarrow\dfrac{x}{20}=\dfrac{y}{24}=\dfrac{z}{21}\)

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

\(\dfrac{x}{20}=\dfrac{y}{24}=\dfrac{z}{21}=\dfrac{x+y-z}{20+24-21}=\dfrac{6}{23}\)(do \(x+y-z=6\))

\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{6}{23}.20=\dfrac{120}{23}\\y=\dfrac{6}{23}.24=\dfrac{144}{23}\\z=\dfrac{6}{23}.21=\dfrac{126}{23}\end{matrix}\right.\)

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Answer 2

Theo đề bài ta có: \(\dfrac{x}{5}=\dfrac{y}{6}\); \(\dfrac{y}{8}=\dfrac{z}{7}\)

\(\Rightarrow\)\(\dfrac{x}{20}=\dfrac{y}{24}\); \(\dfrac{y}{24}=\dfrac{z}{21}\)

\(\Rightarrow\)\(\dfrac{x}{20}=\dfrac{y}{24}=\dfrac{z}{21}\)

Theo t/chất dãy tỉ số bằng nhau, ta có:

\(\dfrac{x}{20}=\dfrac{y}{24}=\dfrac{z}{21}\)\(=\dfrac{x+y-z}{20+24-21}=\dfrac{69}{23}=3\)

Do đó:

\(\dfrac{x}{20}=3\Rightarrow x=60\)

\(\dfrac{y}{24}=3\Rightarrow y=72\)

\(\dfrac{z}{21}=3\Rightarrow z=63\)

Vậy................

Answer 3

Nếu x+y-z=6

Ta có:

\(\dfrac{x}{5}=\dfrac{y}{6}\Rightarrow\dfrac{x}{20}=\dfrac{y}{24}\) (1)

\(\dfrac{y}{8}=\dfrac{z}{7}\Rightarrow\dfrac{y}{24}=\dfrac{z}{21}\)(2)

Từ (1) và (2)

\(\Rightarrow\dfrac{x}{20}=\dfrac{y}{24}=\dfrac{z}{21}\)

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

\(\dfrac{x}{20}=\dfrac{y}{24}=\dfrac{z}{21}=\dfrac{x+y-z}{20+24-21}=\dfrac{6}{23}\) ( Do x+y-z=6)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{20}=\dfrac{6}{23}\\\dfrac{y}{24}=\dfrac{6}{23}\\\dfrac{z}{21}=\dfrac{6}{23}\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{120}{23}\\y=\dfrac{144}{23}\\z=\dfrac{126}{23}\end{matrix}\right.\)

Vậy.....

Answer 4

Nếu x+y-z=69

Ta có:

\(\dfrac{x}{5}=\dfrac{y}{6}\Rightarrow\dfrac{x}{20}=\dfrac{y}{24}\) (1)

\(\dfrac{y}{8}=\dfrac{z}{7}\Rightarrow\dfrac{y}{24}=\dfrac{z}{21}\)(2)

Từ (1) và (2)

\(\Rightarrow\dfrac{x}{20}=\dfrac{y}{24}=\dfrac{z}{21}\)

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

\(\dfrac{x}{20}=\dfrac{y}{24}=\dfrac{z}{21}=\dfrac{x+y-z}{20+24-21}=\dfrac{69}{23}=3\) ( Do x+y-z=69)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{20}=3\\\dfrac{y}{24}=3\\\dfrac{z}{21}=3\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=60\\y=72\\z=63\end{matrix}\right.\)

Vậy.....