\(\dfrac{2x}{3}=\dfrac{3y}{4}=\dfrac{4z}{5}\) và \(x+y+z=49\)
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a) Ta có:
\(x+y+z=49\Rightarrow12x+12y+12z=588\)
Áp dụng tính chất dãy tỉ số bằng nhau, ta có:
\(\dfrac{2x}{3}=\dfrac{3y}{4}=\dfrac{4z}{5}=\dfrac{12x}{18}=\dfrac{12y}{16}=\dfrac{12z}{15}=\dfrac{12x+12y+12z}{18+16+15}=\dfrac{588}{49}=12\)
\(\Rightarrow\left\{{}\begin{matrix}x=12.3:2\\y=12.4:3\\z=12.5:4\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=18\\y=16\\z=15\end{matrix}\right.\)
\(\dfrac{2x}{3}=\dfrac{3y}{4}=\dfrac{4z}{5}=\dfrac{x}{\dfrac{3}{2}}=\dfrac{y}{\dfrac{4}{3}}=\dfrac{z}{\dfrac{5}{4}}\) và \(x+y+z=49\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{2x}{3}=\dfrac{3y}{4}=\dfrac{4z}{5}=\dfrac{x}{\dfrac{3}{2}}=\dfrac{y}{\dfrac{4}{3}}=\dfrac{z}{\dfrac{5}{4}}=\dfrac{x+y+z}{\dfrac{3}{2}+\dfrac{4}{3}+\dfrac{5}{4}}=\dfrac{49}{\dfrac{49}{12}}=12\)\(\Rightarrow\dfrac{x}{\dfrac{3}{2}}=12\Rightarrow x=12.\dfrac{3}{2}=18\)
\(\Rightarrow\dfrac{y}{\dfrac{4}{3}}=12\Rightarrow y=12.\dfrac{4}{3}=16\)
\(\Rightarrow\dfrac{z}{\dfrac{5}{4}}=12\Rightarrow z=12.\dfrac{5}{4}=15\)
Vậy \(\left\{{}\begin{matrix}x=18\\y=16\\z=15\end{matrix}\right.\)
Theo đề bài, ta có:
\(\dfrac{2x}{3}\)=\(\dfrac{3y}{4}\)=\(\dfrac{4z}{5}\)=\(\dfrac{x}{\dfrac{3}{2}}\)=\(\dfrac{y}{\dfrac{4}{3}}\)=\(\dfrac{z}{\dfrac{5}{4}}\) và \(x+y+z=49\)
Áp dụng tính chất dãy tỉ số bằng nhau:
\(\dfrac{2x}{3}\)=\(\dfrac{3y}{4}\)=\(\dfrac{4z}{5}\)=\(\dfrac{x}{\dfrac{3}{2}}\)=\(\dfrac{y}{\dfrac{4}{3}}\)=\(\dfrac{z}{\dfrac{5}{4}}\)=\(\dfrac{x+y+z}{\dfrac{3}{2}+\dfrac{4}{3}+\dfrac{5}{4}}\)=\(\dfrac{49}{\dfrac{18}{12}+\dfrac{16}{12}+\dfrac{15}{12}}\)=\(\dfrac{49}{\dfrac{49}{12}}\)=12
Suy ra: x=12.\(\dfrac{3}{2}\)=18
y=12.\(\dfrac{4}{3}\)=16
z=12.\(\dfrac{5}{4}\)=15
Vậy x=18; y=16; z=15
Ta có :\(\dfrac{2x}{3}=\dfrac{3y}{4}=\dfrac{4z}{5}\Rightarrow\dfrac{2x}{3.12}=\dfrac{3y}{4.12}=\dfrac{4z}{5.12}\)
\(=\dfrac{2x}{36}=\dfrac{3y}{48}=\dfrac{4z}{60}=\dfrac{x}{18}=\dfrac{y}{16}=\dfrac{z}{15}\)và x+y+z=49
\(\Rightarrow\dfrac{x}{18}=\dfrac{y}{16}=\dfrac{z}{15}=\dfrac{x+y+z}{18+16+15}=\dfrac{49}{49}=1\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{18}=1\\\dfrac{y}{16}=1\\\dfrac{z}{15}=1\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=18\\y=16\\z=15\end{matrix}\right.\)
Vậy x=18;y=16;z=15
Ta có: \(\dfrac{2x}{3}=\dfrac{3y}{4}=\dfrac{4z}{5}\Rightarrow\dfrac{12x}{18}=\dfrac{12y}{16}=\dfrac{12z}{15}\)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:
\(\dfrac{12x}{18}=\dfrac{12y}{16}=\dfrac{12z}{15}=\dfrac{12x+12y+12z}{18+16+15}=\dfrac{12.\left(x+y+z\right)}{49}\)
\(=\dfrac{12.49}{49}=12\)
\(\Rightarrow\dfrac{2x}{3}=12\Rightarrow x=18\)
\(\dfrac{3y}{4}=12\Rightarrow y=16\)
\(\dfrac{4z}{5}=12\Rightarrow z=15\)
Vậy \(x=18;y=16;z=15\)
Từ \(\dfrac{2x}{3}=\dfrac{3y}{4}=\dfrac{4z}{5}\Rightarrow\dfrac{x}{\dfrac{3}{2}}=\dfrac{y}{\dfrac{4}{3}}=\dfrac{z}{\dfrac{5}{4}}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có :
\(\dfrac{x}{\dfrac{3}{2}}=\dfrac{y}{\dfrac{4}{3}}=\dfrac{z}{\dfrac{5}{4}}=\dfrac{x+y+z}{\dfrac{3}{2}+\dfrac{4}{3}+\dfrac{5}{4}}=\dfrac{49}{\dfrac{49}{12}}=12\)
⇒\(\dfrac{x}{\dfrac{3}{2}}=12\Rightarrow x=12.\dfrac{3}{2}=18\)
⇒\(\dfrac{y}{\dfrac{4}{3}}=12\Rightarrow y=12.\dfrac{4}{3}=16\)
⇒\(\dfrac{y}{\dfrac{5}{4}}=12\Rightarrow y=12.\dfrac{5}{4}=15\)
Vậy x;y;z lần lượt là 18;16;15
Ta có:
\(\dfrac{2x}{3}=\dfrac{3y}{4}\Rightarrow y=\dfrac{4}{3}.\dfrac{2x}{3}=\dfrac{8x}{9}\)
\(\dfrac{2x}{3}=\dfrac{4z}{5}\Rightarrow z=\dfrac{5}{4}.\dfrac{2x}{3}=\dfrac{10x}{12}=\dfrac{5x}{6}\)
\(\Rightarrow x+y+z=x+\dfrac{8x}{9}+\dfrac{5x}{6}=49\)
Hay \(\left(18+16+15\right).\dfrac{x}{18}=49\).
tức là $x = 18 $
\(\Rightarrow y=16\)
và \(z=15\)
1)
Ta có:
\(2x=3y=4z\Leftrightarrow\dfrac{x}{\dfrac{1}{2}}=\dfrac{y}{\dfrac{1}{3}}=\dfrac{z}{\dfrac{1}{4}}=\dfrac{x-y-z}{\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{4}}=-420\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=-420.\dfrac{1}{2}=-210\\y=-420.\dfrac{1}{3}=-140\\z=-420.\dfrac{1}{4}=-105\end{matrix}\right.\)
Vậy....
ÁP DỤNG TÍNH CHẤT DÃY TỈ SỐ BẰNG NHAU, TA ĐƯỢC :
`(x)/(3)=(y)/(4)=(x+y)/(3+4)=(90)/(7)`
`->` $\begin{cases}x=\dfrac{90}{7}.3=\dfrac{30}{7} \\ y=\dfrac{90}{7}.4=\dfrac{360}{7} \end{cases}$
1)\(\dfrac{x}{5}=\dfrac{y}{3}\) áp dụng...ta đc:
\(\dfrac{x}{5}=\dfrac{y}{3}=\dfrac{x-y}{5-3}=\dfrac{20}{2}=10\)
x=50
y=30
\(\dfrac{2x}{5}=\dfrac{3y}{4}=\dfrac{4z}{5}\)
\(\Rightarrow\dfrac{2}{5}x=\dfrac{3}{4}y=\dfrac{4}{5}z\)
\(\Rightarrow\dfrac{2}{5}x.\dfrac{1}{12}=\dfrac{3}{4}y.\dfrac{1}{12}=\dfrac{4}{5}z.\dfrac{1}{12}\)
\(\Rightarrow\dfrac{x}{30}=\dfrac{y}{16}=\dfrac{z}{15}\)
Đặt \(\dfrac{x}{30}=\dfrac{y}{16}=\dfrac{z}{15}=k\Rightarrow\left\{{}\begin{matrix}x=30k\\y=16k\\z=15k\end{matrix}\right.\). Ta có:
\(x+y+z=49\)
\(\Rightarrow30k+16k+15k=49\)
\(\Rightarrow61k=49\)
\(\Rightarrow k=\dfrac{49}{61}\)
\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{49}{61}.30=\dfrac{1470}{61}\\y=\dfrac{49}{61}.16=\dfrac{784}{61}\\z=\dfrac{49}{61}.15=\dfrac{735}{61}\end{matrix}\right.\)
\(\dfrac{2x}{3}=\dfrac{3y}{4}=\dfrac{4z}{5}\Rightarrow\dfrac{x}{\dfrac{3}{2}}=\dfrac{y}{\dfrac{4}{3}}=\dfrac{z}{\dfrac{5}{4}}\)
Áp dụng t/c dtsbn:
\(\dfrac{x}{\dfrac{3}{2}}=\dfrac{y}{\dfrac{4}{3}}=\dfrac{z}{\dfrac{5}{4}}=\dfrac{x+y+z}{\dfrac{3}{2}+\dfrac{4}{3}+\dfrac{5}{4}}=\dfrac{49}{\dfrac{49}{12}}=12\\ \Rightarrow\left\{{}\begin{matrix}x=12\cdot\dfrac{3}{2}=18\\y=12\cdot\dfrac{4}{3}=16\\z=12\cdot\dfrac{5}{4}=15\end{matrix}\right.\)