Giải:
Theo đề ra, ta có:
\(2x+3y+4z=10\) và \(\dfrac{x}{3}=\dfrac{y}{4}=\dfrac{z}{5}\)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:
\(\dfrac{x}{3}=\dfrac{y}{4}=\dfrac{z}{5}=\dfrac{2x}{6}=\dfrac{3y}{12}=\dfrac{4z}{20}=\dfrac{2x+3y+4z}{6+12+20}=\dfrac{10}{38}=\dfrac{5}{19}\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x}{3}=\dfrac{5}{19}\\\dfrac{y}{4}=\dfrac{5}{19}\\\dfrac{z}{5}=\dfrac{5}{19}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{5}{19}.3\\y=\dfrac{5}{19}.4\\z=\dfrac{5}{19}.5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{15}{19}\\y=\dfrac{20}{19}\\z=\dfrac{25}{19}\end{matrix}\right.\)
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Ta có :
\(\dfrac{x}{3}=\dfrac{y}{4}=\dfrac{z}{5}=\dfrac{2x}{6}=\dfrac{3y}{12}=\dfrac{4z}{20}=\dfrac{2x+3y+4z}{6+12+20}=\dfrac{10}{38}=\dfrac{5}{19}\)
\(\dfrac{x}{3}=\dfrac{5}{19}\Rightarrow x=\dfrac{15}{19}\)
\(\dfrac{y}{4}=\dfrac{5}{19}\Rightarrow y=\dfrac{20}{19}\)
\(\dfrac{z}{5}=\dfrac{5}{19}\Rightarrow z=\dfrac{25}{19}\)
\(\dfrac{x}{3}=\dfrac{y}{4}=\dfrac{z}{5}\)
\(\dfrac{2x}{6}=\dfrac{3y}{12}=\dfrac{4z}{20}=\dfrac{2x+3y+4z}{6+12+20}=\dfrac{10}{38}=\dfrac{5}{19}\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{3}=\dfrac{5}{19}\Rightarrow\dfrac{3.5}{19}=\dfrac{15}{19}\\\dfrac{y}{4}=\dfrac{5}{19}\Rightarrow\dfrac{4.5}{19}=\dfrac{20}{19}\\\dfrac{z}{5}=\dfrac{5}{19}\Rightarrow\dfrac{5.5}{19}=\dfrac{25}{19}\end{matrix}\right.\)
KL:
\(x=\dfrac{15}{19}\\ y=\dfrac{20}{19}\\ z=\dfrac{25}{19}\)
Áp dụng t/c của dãy tỉ sô bằng nhau,ta có:
\(\dfrac{x}{3}=\dfrac{y}{4}=\dfrac{z}{5}=\dfrac{2x+3y+4z}{6+12+20}=\dfrac{10}{38}=\dfrac{5}{19}\)
=>\(x=\dfrac{15}{19}\);\(y=\dfrac{20}{19};z=\dfrac{25}{19}\)
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