$\dfrac{2x}{3}=\dfrac{3y}{4}=\dfrac{4z}{5}$ và $x+y+z=49$ Ta có: $\dfrac{2x}{3}=\dfrac{x}{\dfrac{3}{2}}$ $\dfrac{3y}{4}=\dfrac{y}{\dfrac{4}{3}}$ $\dfrac{4z}{5}=\dfrac{z}{\dfrac{5}{4}}$ Áp dụng tính chất dãy tỉ số bằng nhau, ta có: $\dfrac{x}{\dfrac{3}{2}}=\dfrac{y}{\dfrac{4}{3}}=\dfrac{z}{\dfrac{5}{4}}=\dfrac{x+y+z}{\dfrac{3}{2}+\dfrac{4}{3}+\dfrac{5}{4}}=\dfrac{49}{\dfrac{49}{12}}=12$ $\Rightarrow \left\{\begin{matrix} \dfrac{x}{\dfrac{3}{2}}=12\Rightarrow x=18\\ \dfrac{y}{\dfrac{4}{3}}=12\Rightarrow y=16\\ \dfrac{z}{\dfrac{5}{4}}=12\Rightarrow z=15 \end{matrix}\right.$
$\text{Ta có:}$ `\frac{2x}{3}=\frac{3y}{4}=\frac{4z}{5}` `và` `x+y+z=49`
`⇒\frac{2x}{3}=\frac{2x.6}{3.6}=\frac{12x}{18}`
`⇒\frac{3y}{4}=\frac{3y.4}{4.4}=\frac{12y}{16}`
`⇒\frac{4z}{5}=\frac{4z.3}{5.3}=\frac{12z}{15}`
`⇒\frac{12x}{18}=\frac{12y}{16}=\frac{12z}{15}`
$\text{Áp dụng tính chất dãy tỉ số bằng nhau}$
`⇒\frac{12x+12y+12z}{18+16+15}=\frac{12(x+y+z)}{49}`
`=\frac{12.49}{49}=12`
`⇒\frac{2x}{3}=12⇒x=18`
`⇒\frac{3y}{4}=24`
`⇒\frac{4z}{5}=30`
Đáp án: $x=18; y=16; z=15$
Giải thích các bước giải:
$\dfrac{2x}{3}=\dfrac{3y}{4}=\dfrac{4z}{5}$ và $x+y+z=49$
Ta có:
$\dfrac{2x}{3}=\dfrac{x}{\dfrac{3}{2}}$
$\dfrac{3y}{4}=\dfrac{y}{\dfrac{4}{3}}$
$\dfrac{4z}{5}=\dfrac{z}{\dfrac{5}{4}}$
Áp dụng tính chất dãy tỉ số bằng nhau, ta có:
$\dfrac{x}{\dfrac{3}{2}}=\dfrac{y}{\dfrac{4}{3}}=\dfrac{z}{\dfrac{5}{4}}=\dfrac{x+y+z}{\dfrac{3}{2}+\dfrac{4}{3}+\dfrac{5}{4}}=\dfrac{49}{\dfrac{49}{12}}=12$
$\Rightarrow \left\{\begin{matrix}
\dfrac{x}{\dfrac{3}{2}}=12\Rightarrow x=18\\
\dfrac{y}{\dfrac{4}{3}}=12\Rightarrow y=16\\
\dfrac{z}{\dfrac{5}{4}}=12\Rightarrow z=15
\end{matrix}\right.$