Question

Tìm 3 số : x;y;z. Biết rằng :

10x = 2y = 5z và x+y-z = 3

Answer 1

Ta có : \(10x=2y\Rightarrow\dfrac{x}{2}=\dfrac{y}{10}\)(1);

\(2y=5z\Rightarrow\dfrac{y}{5}=\dfrac{z}{2}\Rightarrow\dfrac{y}{10}=\dfrac{z}{4}\)(2);

Từ (1) và (2)=> \(\dfrac{x}{2}=\dfrac{y}{10}=\dfrac{z}{4}\)và \(x+y+z=3\)

Theo tính chất dãu tỉ số bằng nhau , ta có :

\(\dfrac{x}{2}=\dfrac{y}{10}=\dfrac{z}{4}=\dfrac{x+y-z}{2+10-4}=\dfrac{3}{8}\)

\(\Rightarrow x=\dfrac{3}{8}.2=\dfrac{3}{4}\)

\(\Rightarrow y=\dfrac{3}{8}.10=\dfrac{15}{4}\)

\(\Rightarrow z=\dfrac{3}{8}.4=\dfrac{3}{2}\)

Tik mik nha !!!

Answer 2

- Theo đề bài ta có:

10x = 2y = 5z

<=> \(\dfrac{10x}{10}=\dfrac{2y}{10}=\dfrac{5z}{10}\)

<=> \(\dfrac{x}{1}=\dfrac{y}{5}=\dfrac{z}{2}\)

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

\(\dfrac{x}{1}=\dfrac{y}{5}=\dfrac{z}{2}\)=\(\dfrac{x+y-z}{1+5-2}=\dfrac{3}{4}\)

- Vậy x = \(\dfrac{3}{4}\cdot1=\dfrac{3}{4}\)

y = \(\dfrac{3}{4}\cdot5=\dfrac{15}{4}\)

z = \(\dfrac{3}{4}\cdot2\)=\(\dfrac{6}{4}=\dfrac{3}{2}\)

Answer 3

Bài này có thể giải = 2 cách nhé:

Đặt: \(10x=2y=5z=k\)

\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{k}{10}\\y=\dfrac{k}{2}\\z=\dfrac{k}{5}\end{matrix}\right.\)

\(\Rightarrow\dfrac{k}{10}+\dfrac{k}{2}-\dfrac{k}{5}=3\)

\(\Rightarrow\dfrac{k}{10}+\dfrac{5k}{10}-\dfrac{2k}{10}=3\)

\(\Rightarrow\dfrac{4k}{10}=3\)

\(\Rightarrow4k=30\Rightarrow k=\dfrac{30}{4}\)

\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{30}{4}:10=\dfrac{3}{4}\\y=\dfrac{30}{4}:2=\dfrac{15}{4}\\z=\dfrac{30}{4}:5=\dfrac{3}{2}\end{matrix}\right.\)

Answer 4

Hơi rối nhá

Ta có:

\(10x=2y=5z\Rightarrow\dfrac{x}{\dfrac{1}{10}}=\dfrac{y}{\dfrac{1}{2}}=\dfrac{z}{\dfrac{1}{5}}\)

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

\(\dfrac{x}{\dfrac{1}{10}}=\dfrac{y}{\dfrac{1}{2}}=\dfrac{z}{\dfrac{1}{5}}=\dfrac{x+y-z}{\dfrac{1}{10}+\dfrac{1}{2}-\dfrac{1}{5}}=\dfrac{3}{\dfrac{2}{5}}=7,5\) (vì x+y-z=3)

\(\Rightarrow\left\{{}\begin{matrix}x=7,5.\dfrac{1}{10}=0,75\\y=7,5.\dfrac{1}{2}=3,75\\z=7,5.\dfrac{1}{5}=1,5\end{matrix}\right.\)

Vậy....

Answer 5

Đặt 10x=2y=5z=k

=> \(x=\dfrac{k}{10};y=\dfrac{k}{2};z=\dfrac{k}{5}\)

Ta có: x+y-z=3

\(\Rightarrow\dfrac{k}{10}+\dfrac{k}{2}-\dfrac{k}{5}=3\)

\(\Leftrightarrow\dfrac{k+5k-2k}{10}=3\)

\(\Leftrightarrow4k=30\)

\(\Leftrightarrow k=\dfrac{15}{2}\)

\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{3}{4}\\y=\dfrac{15}{4}\\z=\dfrac{3}{2}\end{matrix}\right.\)

Vậy ...