Question

Answer 1

\(x-\left(y+z\right)=21\Rightarrow x-y-z=21\)

\(3x=4y=5z\Rightarrow\dfrac{x}{\dfrac{1}{3}}=\dfrac{y}{\dfrac{1}{4}}=\dfrac{z}{\dfrac{1}{5}}\)

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

\(\dfrac{x}{\dfrac{1}{3}}=\dfrac{y}{\dfrac{1}{4}}=\dfrac{z}{\dfrac{1}{5}}=\dfrac{x-y-z}{\dfrac{1}{3}-\dfrac{1}{4}-\dfrac{1}{5}}=\dfrac{21}{-\dfrac{1}{15}}=-315\)

\(\dfrac{x}{\dfrac{1}{3}}=-315\text{​​}\text{​​}\Rightarrow x=-105\\ \dfrac{y}{\dfrac{1}{4}}=-315\Rightarrow y=-\dfrac{315}{4}\\ \dfrac{z}{\dfrac{1}{5}}=-315\Rightarrow-63\)

Answer 2

$3x=4y\to \dfrac x 4=\dfrac  y 3$

$4y=5z\to \dfrac y 5=\dfrac z 4$

$\to \dfrac{x}{20}=\dfrac{y}{15}=\dfrac{z}{12}$

Áp dụng tính chất dãy tỉ số bằng nhau

$\dfrac{x}{20}=\dfrac{y}{15}=\dfrac{z}{12}=\dfrac{x-(y+z)}{20-(15+12)}=\dfrac{-21}{-7}=3$

$\to \begin{cases}x=60\\y=45\\z=36\end{cases}$

Vậy $(x;y;z)=(60;45;36)$

Answer 3

Ta có: \(3x=4y=5z\)

nên \(\dfrac{x}{\dfrac{1}{3}}=\dfrac{y}{\dfrac{1}{4}}=\dfrac{z}{\dfrac{1}{5}}\)

mà x-y-z=-21

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

\(\dfrac{x}{\dfrac{1}{3}}=\dfrac{y}{\dfrac{1}{4}}=\dfrac{z}{\dfrac{1}{5}}=\dfrac{x-y-z}{\dfrac{1}{3}-\dfrac{1}{4}-\dfrac{1}{5}}=\dfrac{-21}{-\dfrac{7}{60}}=180\)

Do đó: x=60; y=45; z=36