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1. \(\dfrac{x}{19}=\dfrac{y}{21};2x-y=34\)
Có: \(\dfrac{x}{19}=\dfrac{y}{21}\)
=> \(\dfrac{2x}{38}=\dfrac{y}{21}\)
Theo tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{2x}{38}=\dfrac{y}{21}=\dfrac{2x-y}{38-21}=\dfrac{34}{17}=2\)
=> \(\dfrac{x}{19}=2=>x=2.19=38\)
=> \(\dfrac{y}{21}=2=>y=2.21=42\)
Vậy x= 38 ; y= 42
2. \(\dfrac{x}{15}=\dfrac{y}{20}=\dfrac{z}{28}\);\(2x+3y-z=186\)
Có: \(\dfrac{x}{15}=\dfrac{y}{20}=\dfrac{z}{28}\)
=> \(\dfrac{2x}{30}=\dfrac{3y}{60}=\dfrac{z}{28}=\dfrac{2x+3y-z}{30+60-28}=\dfrac{186}{62}=3\)
=> \(\dfrac{x}{15}=3=>x=3.15=45\)
=>\(\dfrac{y}{20}=3=>y=3.20=60\)
=> \(\dfrac{z}{28}=3=>z=3.28=84\)
Vậy x=45;y=60;z=84
1) \(\dfrac{x}{19}=\dfrac{y}{21}\) và 2x -y =34
Từ \(\dfrac{x}{19}=\dfrac{y}{21}=>\dfrac{2x}{38}=\dfrac{y}{21}\)
Áp dụng tính chất của dãy tỉ số bằng nhau,ta có:
\(\dfrac{2x}{38}=\dfrac{y}{21}=\dfrac{2x-y}{38-21}=\dfrac{34}{17}=2\)
=>\(\dfrac{2x}{38}=2=>2x=2.38=>2x=76=>x=76:2=>x=38\)
=>\(\dfrac{y}{21}=2=>y=2.21=>y=42\)
Vậy x=38; y=42
2)\(\dfrac{x}{15}=\dfrac{y}{20}=\dfrac{z}{28}\)và 2x+3y-z=186
Từ \(\dfrac{x}{15}=\dfrac{y}{20}=\dfrac{z}{28}=>\dfrac{2x}{30}=\dfrac{3y}{60}=\dfrac{z}{28}\)
Áp dụng t/c của dãy tỉ số bằng nhau,ta có:
\(\dfrac{2x}{30}=\dfrac{3y}{60}=\dfrac{z}{28}=\dfrac{2x+3y-z}{30+60-28}=\dfrac{186}{62}=3\)
=>\(\dfrac{2x}{30}=3=>2x=3.30=>2x=90=>x=90:2=>x=45\)
=>\(\dfrac{3y}{60}=3=>3y=3.60=>3y=180=>y=180:3=>y=60\)
=>\(\dfrac{z}{28}=3=>z=3.28=>z=84\)
Vậy x=45; y=60; z=84
3)\(\dfrac{x}{3}=\dfrac{y}{4}\) và\(\dfrac{y}{5}=\dfrac{z}{7}\)và 2x+3y-z=372
Từ\(\dfrac{x}{3}=\dfrac{y}{4}=>\dfrac{x}{15}=\dfrac{y}{20}\)
\(\dfrac{y}{5}=\dfrac{z}{7}=>\dfrac{y}{20}=\dfrac{z}{28}\)
=>\(\dfrac{x}{15}=\dfrac{y}{20}=\dfrac{z}{28}=>\dfrac{2x}{30}=\dfrac{3y}{60}=\dfrac{z}{28}\)
Áp dụng t/c của dãy tỉ số bằng nhau,ta có:
\(\dfrac{2x}{30}=\dfrac{3y}{60}=\dfrac{z}{28}=\dfrac{2x+3y-z}{30+60-28}=\dfrac{372}{62}=6\)
=>\(\dfrac{2x}{30}=6=>2x=6.30=>2x=180=>x=180:2=>x=90\)
=>\(\dfrac{3y}{60}=6=>3y=6.60=>3y=360=>y=360:3=>y=120\)
=>\(\dfrac{z}{28}=6=>z=6.28=>z=148\)
Vậy x=90; y=120; z=148
\(1,\dfrac{x}{2}=\dfrac{y}{5}=\dfrac{x+y}{2+5}=\dfrac{21}{7}=3\\ \Rightarrow\left\{{}\begin{matrix}x=6\\y=15\end{matrix}\right.\\ 2,7x=3y\Rightarrow\dfrac{x}{3}=\dfrac{y}{7}=\dfrac{x-y}{3-7}=\dfrac{16}{-4}=-4\\ \Rightarrow\left\{{}\begin{matrix}x=-12\\y=-28\end{matrix}\right.\\ 3,\dfrac{x}{5}=\dfrac{y}{6}=\dfrac{z}{7}=\dfrac{x-y-z}{5-6-7}=\dfrac{36}{-8}=-\dfrac{9}{2}\\ \Rightarrow\left\{{}\begin{matrix}x=-\dfrac{45}{2}\\y=-27\\z=-\dfrac{63}{2}\end{matrix}\right.\\ 4,x:y:z=3:5:7\Rightarrow\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{z}{7}=\dfrac{2x+3y-z}{6+15-7}=\dfrac{-14}{14}=-1\\ \Rightarrow\left\{{}\begin{matrix}x=-3\\y=-5\\z=-7\end{matrix}\right.\)
3. Áp dụng tính chất dãy tỉ số bằng nhau, ta có:
\(\dfrac{x}{5}=\dfrac{y}{6}=\dfrac{z}{7}=\dfrac{x-y-z}{5-6-7}=\dfrac{36}{-8}=\dfrac{-9}{2}\)
\(x=\dfrac{-45}{2}\)
\(y=-27\)
\(z=\dfrac{-63}{2}\)
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 ...
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: ...
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:
\(\left\{{}\begin{matrix}\dfrac{x}{15}=2\\\dfrac{y}{20}=2\\\dfrac{z}{28}=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=30\\y=40\\z=56\end{matrix}\right.\)
\(\dfrac{x}{3}=\dfrac{x}{4}\Leftrightarrow\dfrac{x}{15}=\dfrac{y}{20}\\ \dfrac{y}{5}=\dfrac{z}{7}\Leftrightarrow\dfrac{y}{20}=\dfrac{z}{28}\)
\(\Rightarrow\dfrac{x}{15}=\dfrac{y}{20}=\dfrac{z}{28}=\dfrac{2x+3y-z}{2.15+3.20-28}=\dfrac{125}{62}=2\)
\(\Rightarrow\left\{{}\begin{matrix}x=15.2=30\\y=20.2=40\\z=28.2=56\end{matrix}\right.\)
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....
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
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:
\(\dfrac{x-y+z}{9-7+3}=\dfrac{-15}{5}=-3\)
=> \(\left\{{}\begin{matrix}x=-27\\y=-21\\z=-9\end{matrix}\right.\)
\(\dfrac{x}{3}=\dfrac{y}{4}\)⇒\(\dfrac{x}{15}=\dfrac{y}{20}\)
\(\dfrac{y}{5}=\dfrac{z}{7}\)⇒\(\dfrac{y}{20}=\dfrac{z}{28}\)
⇒\(\dfrac{x}{15}=\dfrac{y}{20}=\dfrac{z}{28}\)⇒\(\dfrac{2x}{30}=\dfrac{3y}{60}=\dfrac{z}{28}\)
Áp dụng tính chất dãy tỉ số bằng nhau, ta có:
\(\dfrac{2x}{30}=\dfrac{3y}{60}=\dfrac{z}{28}=\dfrac{2x+3y-z}{30+60-28}=\dfrac{372}{62}=6\)
⇒\(\left\{{}\begin{matrix}x=6.15=90\\y=6.20=120\\z=6.28=168\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\dfrac{x}{3}=\dfrac{y}{4}\\\dfrac{y}{5}=\dfrac{z}{7}\end{matrix}\right.\)\(\Rightarrow\dfrac{x}{15}=\dfrac{y}{20}=\dfrac{z}{28}\)
Áp dụng t/c dtsbn:
\(\dfrac{x}{15}=\dfrac{y}{20}=\dfrac{z}{28}=\dfrac{2x}{30}=\dfrac{3y}{60}=\dfrac{2x+3y-z}{30+60-28}=\dfrac{372}{62}=6\)
\(\Rightarrow\left\{{}\begin{matrix}x=6.15=90\\y=6.20=120\\z=6.28=168\end{matrix}\right.\)