\(\dfrac{x}{2}=\dfrac{y}{3},\dfrac{y}{5}=\dfrac{z}{7}\) và 2x+3y+z=172
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\(1,\dfrac{x}{2}=\dfrac{y}{5}=\dfrac{x+y}{2+5}=\dfrac{21}{7}=3\\ \Rightarrow\left\{{}\begin{matrix}x=6\\y=15\end{matrix}\right.\\ 2,7x=3y\Rightarrow\dfrac{x}{3}=\dfrac{y}{7}=\dfrac{x-y}{3-7}=\dfrac{16}{-4}=-4\\ \Rightarrow\left\{{}\begin{matrix}x=-12\\y=-28\end{matrix}\right.\\ 3,\dfrac{x}{5}=\dfrac{y}{6}=\dfrac{z}{7}=\dfrac{x-y-z}{5-6-7}=\dfrac{36}{-8}=-\dfrac{9}{2}\\ \Rightarrow\left\{{}\begin{matrix}x=-\dfrac{45}{2}\\y=-27\\z=-\dfrac{63}{2}\end{matrix}\right.\\ 4,x:y:z=3:5:7\Rightarrow\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{z}{7}=\dfrac{2x+3y-z}{6+15-7}=\dfrac{-14}{14}=-1\\ \Rightarrow\left\{{}\begin{matrix}x=-3\\y=-5\\z=-7\end{matrix}\right.\)
3. Áp dụng tính chất dãy tỉ số bằng nhau, ta có:
\(\dfrac{x}{5}=\dfrac{y}{6}=\dfrac{z}{7}=\dfrac{x-y-z}{5-6-7}=\dfrac{36}{-8}=\dfrac{-9}{2}\)
\(x=\dfrac{-45}{2}\)
\(y=-27\)
\(z=\dfrac{-63}{2}\)
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a,Áp sụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{z}{7}=\dfrac{3x-2z}{9-14}=\dfrac{15}{-5}=-3\\\Rightarrow x=-3.3=-9\\ \Rightarrow y=-3.5=-15\\ \Rightarrow z=-3.7=-21 \)
a) Ta có: \(\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{z}{7}=\dfrac{3x}{9}=\dfrac{2z}{14}=\dfrac{3x-2z}{9-14}=\dfrac{15}{-5}=-3\) (Vì 3x-2z=15)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{3}=-3\\\dfrac{y}{5}=-3\\\dfrac{z}{7}=-3\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=-9\\y=-15\\z=-21\end{matrix}\right.\)
Vậy ...
b) Ta có: \(\dfrac{x}{5}=\dfrac{y}{3}=\dfrac{z}{2}=\dfrac{2x}{10}=\dfrac{3y}{9}=\dfrac{2x-3y}{10-9}=\dfrac{100}{1}=100\) (Vì 2x-3y=100)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{5}=100\\\dfrac{y}{3}=100\\\dfrac{z}{2}=100\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=500\\y=300\\z=200\end{matrix}\right.\)
Vậy ...
c) Ta có: \(\dfrac{x}{-3}=\dfrac{y}{-5}=\dfrac{z}{-4}=\dfrac{3z}{-12}=\dfrac{2x}{-6}=\dfrac{3z-2x}{\left(-12\right)-\left(-6\right)}=\dfrac{36}{-18}=-2\) (Vì 3z-2x=36)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{-3}=-2\\\dfrac{y}{-5}=-2\\\dfrac{z}{-4}=-2\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=6\\y=10\\z=8\end{matrix}\right.\)
Vậy ...
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4: Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{x}{8}=\dfrac{y}{12}=\dfrac{z}{15}=\dfrac{x-y-z}{8-12-15}=\dfrac{38}{-19}=-2\)
Do đó: x=-16; y=-24; z=-30
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7) vì \(\dfrac{x}{5}\)=\(\dfrac{y}{6}\)=\(\dfrac{z}{7}\)và x-y+z=36
Nên theo tính chất của dãy tỉ số bằng nhau ta có:
\(\dfrac{x}{5}\)=\(\dfrac{y}{6}\)=\(\dfrac{z}{7}\)=\(\dfrac{x-y+z}{5-6+7}\)=\(\dfrac{36}{6}\)=6
\(\Rightarrow\)x=6.5=30
y=6.6=36
z=6.7=42
vậy x=30,y=36,z=42
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Ta có: \(\dfrac{x}{3}=\dfrac{y}{4}\Rightarrow\dfrac{x}{15}=\dfrac{y}{20}\)
\(\dfrac{y}{5}=\dfrac{z}{7}\Rightarrow\dfrac{y}{20}=\dfrac{z}{28}\)
\(\Rightarrow\dfrac{x}{15}=\dfrac{y}{20}=\dfrac{z}{28}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{x}{15}=\dfrac{y}{20}=\dfrac{y}{28}=\dfrac{2x+3y-z}{15\cdot2+3\cdot20-28}=\dfrac{186}{62}=3\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{15}=3\Rightarrow x=45\\\dfrac{y}{20}=3\Rightarrow y=60\\\dfrac{z}{28}=3\Rightarrow z=84\end{matrix}\right.\)
Vậy: ...
�3=�4;�5=�7⇒�15=�20;�20=�283x=4y;5y=7z⇒15x=20y;20y=28z
Áp dụng tính chất dãy tỉ số bằng nhau :
�15=�20=�28=2�+3�−�30+60−28=18662=315x=20y=28z=30+60−282x+3y−z=62186=3
⇒�15=3.15=45⇒15x=3.15=45
⇒�20=3.20=60⇒20y=3.20=60
⇒�28=3.28=84⇒28z=3.28=84
Vậy x = 45; y= 60; z = 84
![](https://rs.olm.vn/images/avt/0.png?1311)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{z}{7}=\dfrac{2x+3y-z}{2\cdot3+3\cdot5-7}=\dfrac{373}{14}\)
Do đó: \(\left\{{}\begin{matrix}x=\dfrac{1119}{14}\\y=\dfrac{1865}{14}\\z=\dfrac{373}{2}\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
1. Ta có: \(\dfrac{x}{-7}=\dfrac{y}{4}\Rightarrow\dfrac{2x}{-14}=\dfrac{3y}{12}\)
Áp dụng tính chất dãy tỉ số bằng nhau, ta được:
\(\dfrac{2x-3y}{-14-12}=\dfrac{-78}{-26}=3\)
=> \(\left\{{}\begin{matrix}x=-21\\y=12\end{matrix}\right.\)
2. Ta có:
- \(\dfrac{x}{y}=\dfrac{9}{7}\Rightarrow\dfrac{x}{9}=\dfrac{y}{7}\)
- \(\dfrac{y}{z}=\dfrac{7}{3}\Rightarrow\dfrac{y}{7}=\dfrac{z}{3}\)
=> \(\dfrac{x}{9}=\dfrac{y}{7}=\dfrac{z}{3}\)
Áp dụng tính chất dãy tỉ số bằng nhau, ta được:
\(\dfrac{x-y+z}{9-7+3}=\dfrac{-15}{5}=-3\)
=> \(\left\{{}\begin{matrix}x=-27\\y=-21\\z=-9\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: \(\dfrac{x}{3}=\dfrac{y}{4}\)
nên \(\dfrac{x}{15}=\dfrac{y}{20}\)(1)
Ta có: \(\dfrac{y}{5}=\dfrac{z}{7}\)
nên \(\dfrac{y}{20}=\dfrac{z}{28}\)(2)
Từ (1) và (2) suy ra \(\dfrac{x}{15}=\dfrac{y}{20}=\dfrac{z}{28}\)
\(\Leftrightarrow\dfrac{2x}{30}=\dfrac{3y}{60}=\dfrac{z}{28}\)
mà 2x+3y-z=124
nên Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{2x}{30}=\dfrac{3y}{60}=\dfrac{z}{28}=\dfrac{2x+3y-z}{30+60-28}=\dfrac{124}{62}=2\)
Do đó:
\(\left\{{}\begin{matrix}\dfrac{x}{15}=2\\\dfrac{y}{20}=2\\\dfrac{z}{28}=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=30\\y=40\\z=56\end{matrix}\right.\)
\(\dfrac{x}{3}=\dfrac{x}{4}\Leftrightarrow\dfrac{x}{15}=\dfrac{y}{20}\\ \dfrac{y}{5}=\dfrac{z}{7}\Leftrightarrow\dfrac{y}{20}=\dfrac{z}{28}\)
\(\Rightarrow\dfrac{x}{15}=\dfrac{y}{20}=\dfrac{z}{28}=\dfrac{2x+3y-z}{2.15+3.20-28}=\dfrac{125}{62}=2\)
\(\Rightarrow\left\{{}\begin{matrix}x=15.2=30\\y=20.2=40\\z=28.2=56\end{matrix}\right.\)
Theo đề bài, ta có :
\(\dfrac{x}{2}=\dfrac{y}{3}\) \(\Rightarrow\) \(\dfrac{x}{10}=\dfrac{y}{15}\) \(\Rightarrow\) \(\dfrac{2x}{20}=\dfrac{3y}{45}\) (1)
\(\dfrac{y}{5}=\dfrac{z}{7}\Rightarrow\dfrac{y}{15}=\dfrac{z}{21}\Rightarrow\dfrac{3y}{45}=\dfrac{z}{21}\) (2)
Từ (1) và (2) \(\Rightarrow\) \(\dfrac{2x}{20}=\dfrac{3y}{45}\)\(=\dfrac{z}{21}\)
Mà 2x + 3y + z = 172
Theo tính chất dãy tỉ số bằng nhau, ta có :
\(\dfrac{2x}{20}=\dfrac{3y}{45}\)\(=\dfrac{z}{21}\) \(=\dfrac{2x+3y+z}{20+45+21}\) = 2
Do đó : \(\dfrac{2x}{20}=2\Rightarrow x=2.20:2=20\)
\(\dfrac{3y}{45}=2\Rightarrow y=2.45:3=30\)
\(\dfrac{z}{21}=2\Rightarrow z=2.21=42\)
Vật x = 20 ; y = 30 ; z = 42
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