Question

Tìm x,y,z

\(\dfrac{2x}{3}=\dfrac{3y}{4}=\dfrac{4z}{5}\) và x+y+z = 147

Answer 1

Giải:

Ta có:

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

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

\(\Leftrightarrow\dfrac{x}{\dfrac{18}{12}}=\dfrac{y}{\dfrac{16}{12}}=\dfrac{z}{\dfrac{15}{12}}\)

\(\Leftrightarrow\dfrac{x}{18}=\dfrac{y}{16}=\dfrac{z}{15}\)

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

\(\dfrac{x}{18}=\dfrac{y}{16}=\dfrac{z}{15}=\dfrac{x+y+z}{18+16+15}=\dfrac{147}{49}=3\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x}{18}=3\\\dfrac{y}{16}=3\\\dfrac{z}{15}=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=54\\y=48\\z=45\end{matrix}\right.\)

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Answer 2

Ta có:

$\frac{2x}{3}=\frac{3y}{4}=\frac{4z}{5}$

$\iff \frac{x}{\frac{3}{2}}=\frac{y}{\frac{4}{3}}=\frac{z}{\frac{5}{4}}$

$\iff \frac{x}{\frac{18}{12}}=\frac{y}{\frac{16}{12}}=\frac{z}{\frac{15}{12}}$

$\iff \frac{x}{18}=\frac{y}{16}=\frac{z}{15}$

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

$\frac{x}{18}=\frac{y}{16}=\frac{z}{15}=\frac{x+y+z}{18+16+15}=\frac{147}{49}=3$

$\iff \{\begin{array}{c}\frac{x}{18}=3\\ \frac{y}{16}=3\\ \frac{z}{15}=3\end{array}\iff \{\begin{array}{c}x=54\\ y=48\\ z=45\end{array}$