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a) Giải
Vì \(5x=2y=3z\)
\(\Rightarrow\dfrac{5x}{30}=\dfrac{2y}{30}=\dfrac{3z}{30}\)
\(\Rightarrow\dfrac{x}{6}=\dfrac{y}{15}=\dfrac{z}{10}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{x}{6}=\dfrac{y}{15}=\dfrac{z}{10}=\dfrac{x+y-z}{6+15-10}=\dfrac{33}{11}=3\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{6}=3\Rightarrow x=18\\\dfrac{y}{15}=3\Rightarrow y=45\\\dfrac{z}{10}=3\Rightarrow z=30\end{matrix}\right.\)
Vậy \(x=18,\) \(y=45\) hoặc \(z=30.\)
c) Giải
(Vì mk bt bạn bấm nhầm nên đề bị sai, mk sửa 7 \(\rightarrow\) y do trên bàn phím, 7 với y ở vị trí gần nhau mà 2 với y ở cách xa nhau nên sửa như vậy nhé)
Vì \(\dfrac{x-1}{2}=\dfrac{y-2}{3}=\dfrac{z-3}{4}\)
\(\Rightarrow\dfrac{x-1}{2}=\dfrac{y-2}{3}=\dfrac{z-3}{4}=\dfrac{2y-4}{6}=\dfrac{3z-9}{12}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{x-1}{2}=\dfrac{y-2}{3}=\dfrac{z-3}{4}=\dfrac{\left(x-1\right)-\left(2y-4\right)+\left(3z-9\right)}{4-6+12}=\dfrac{x-1-2y+4+3z-9}{10}\)
\(=\dfrac{\left(x-2y+3z\right)-\left(1-4+9\right)}{10}=\dfrac{14-6}{10}=\dfrac{4}{5}\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x-1}{2}=\dfrac{4}{5}\Rightarrow x=\dfrac{13}{5}\\\dfrac{y-2}{3}=\dfrac{4}{5}\Rightarrow y=\dfrac{22}{5}\\\dfrac{z-3}{4}=\dfrac{4}{5}\Rightarrow z=\dfrac{31}{5}\end{matrix}\right.\)
Vậy \(x=\dfrac{13}{5},\) \(y=\dfrac{22}{5}\) và \(z=\dfrac{31}{5}.\)
c) Giải
Đặt \(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{5}=k\Rightarrow\left\{{}\begin{matrix}x=2k\\y=3k\\z=5k\end{matrix}\right.\)
Mà \(x^2+2y^2-z^2=-12\)
\(\Rightarrow\left(2k\right)^2+2\left(3k\right)^2-\left(5k\right)^2=-12\)
\(\Rightarrow4.k^2+18.k^2-25.k^2=-12\)
\(\Rightarrow\left(-3\right)k^2=-12\)
\(\Rightarrow k^2=4\)
\(\Rightarrow k=\pm2\)
\(\circledast k=-2\Rightarrow\left\{{}\begin{matrix}x=-4\\y=-6\\z=-10\end{matrix}\right.\)
\(\circledast k=2\Rightarrow\left\{{}\begin{matrix}x=4\\y=6\\z=10\end{matrix}\right.\)
Vậy \(\left[{}\begin{matrix}x=-4;y=-6;z=-10\\x=4;y=6;z=10\end{matrix}\right..\)
Cho \(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}\)
\(A=\dfrac{x+y-z}{x-y+z};B=\dfrac{2x+3y+z}{x-2y-3z}\)
\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}=k\)
\(\Rightarrow x=2k;y=3k;z=4k\)
sau đó bạn tự thay vào A và B r tính nhá
Đặt:
\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}=k\)
\(\Rightarrow\left\{{}\begin{matrix}x=2k\\y=3k\\z=4k\end{matrix}\right.\)
\(\Rightarrow A=\dfrac{2k+3k-4k}{2k-3k+4k}=\dfrac{k}{3k}=\dfrac{1}{3}\)
\(\Rightarrow B=\dfrac{2.2k+3.3k+4k}{2k-2.3k-3.4k}=\dfrac{4k+9k+4k}{2k-6k-12k}=\dfrac{17k}{-16k}=\dfrac{17}{-16}\)
1/ a, Ta có :
\(x-2y+3z=35\)
\(\dfrac{x}{3}=\dfrac{y}{4}=\dfrac{z}{5}\)
\(\Leftrightarrow\dfrac{x}{3}=\dfrac{2y}{8}=\dfrac{3z}{15}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có :
\(\dfrac{x}{3}=\dfrac{2y}{8}=\dfrac{3z}{15}=\dfrac{x-2y+3z}{3-8+15}=\dfrac{35}{10}=\dfrac{7}{2}\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x}{3}=\dfrac{7}{2}\Leftrightarrow x=\dfrac{21}{2}\\\dfrac{x}{4}=\dfrac{7}{2}\Leftrightarrow y=14\\\dfrac{z}{5}=\dfrac{7}{2}\Leftrightarrow z=\dfrac{35}{2}\end{matrix}\right.\)
Vậy ..
a) Ta có: \(6x=4y=3z\Rightarrow\dfrac{6x}{12}=\dfrac{4y}{12}=\dfrac{3z}{12}\Rightarrow\dfrac{x}{2}=\dfrac{2y}{6}=\dfrac{3z}{12}\)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:
\(\dfrac{x}{2}=\dfrac{2y}{6}=\dfrac{3z}{12}=\dfrac{x+2y-3z}{2+6-12}=\dfrac{-2}{-4}=\dfrac{1}{2}.\)
Với: \(\dfrac{x}{2}=\dfrac{1}{2}\Rightarrow x=1.\)
\(\dfrac{2y}{6}=\dfrac{y}{3}=\dfrac{1}{2}\Rightarrow y=\dfrac{1}{2}.3=\dfrac{3}{2}.\)
\(\dfrac{3z}{12}=\dfrac{z}{4}=\dfrac{1}{2}\Rightarrow z=\dfrac{1}{2}.4=\dfrac{4}{2}=2.\)
Vậy: \(x=1;y=\dfrac{3}{2};z=2.\)
a)\(\dfrac{x}{8}=\dfrac{y}{5}=\dfrac{z}{12}\Leftrightarrow\dfrac{-x}{-8}=\dfrac{y}{5}=\dfrac{z}{12}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{-x}{-8}=\dfrac{y}{5}=\dfrac{z}{12}=\dfrac{-x+y+z}{-8+5+12}=\dfrac{60}{9}=\dfrac{20}{3}\)
\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{20}{3}.8=\dfrac{160}{3}\\y=\dfrac{20}{3}.5=\dfrac{100}{3}\\z=\dfrac{20}{3}.12=80\end{matrix}\right.\)
b) \(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}\Leftrightarrow\dfrac{x}{2}=\dfrac{2y}{6}=\dfrac{3z}{12}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{x}{2}=\dfrac{2y}{6}=\dfrac{3z}{12}=\dfrac{x+2y-3z}{2+6-12}=\dfrac{-20}{-4}=5\)
\(\Rightarrow\left\{{}\begin{matrix}x=5.2=10\\y=5.3=15\\z=5.4=20\end{matrix}\right.\)
c) \(\left\{{}\begin{matrix}4x=3y\\7y=5z\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x}{3}=\dfrac{y}{4}\\\dfrac{y}{5}=\dfrac{z}{7}\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{15}=\dfrac{y}{20}\\\dfrac{y}{20}=\dfrac{z}{28}\end{matrix}\right.\) \(\Leftrightarrow\dfrac{x}{15}=\dfrac{y}{20}=\dfrac{z}{28}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{x}{15}=\dfrac{y}{20}=\dfrac{z}{28}=\dfrac{x-y+z}{15-20+28}=\dfrac{-46}{23}=-2\)
\(\Rightarrow\left\{{}\begin{matrix}x=-2.15=-30\\y=-2.20=-40\\z=-2.28=-56\end{matrix}\right.\)
\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}\Rightarrow\left\{{}\begin{matrix}x=\dfrac{2y}{3}\\x=\dfrac{z}{2}\\y=\dfrac{3z}{4}\end{matrix}\right.\)
ta có \(\dfrac{3x-2y}{4}=\dfrac{2z-4x}{3}=\dfrac{4y-3z}{2}\)
\(\Leftrightarrow\dfrac{3\dfrac{2y}{3}-2y}{4}=\dfrac{2z-4\dfrac{z}{2}}{3}=\dfrac{4\dfrac{3z}{4}-3z}{2}\)
\(\Leftrightarrow0=0=0\)( luôn đúng)
suy ra Đpcm
\(\dfrac{3x-2y}{4}=\dfrac{2z-4x}{3}=\dfrac{4y-3z}{2}\)
\(\Rightarrow\dfrac{4\left(3x-2y\right)}{16}=\dfrac{3\left(2z-4x\right)}{9}=\dfrac{2\left(4y-3z\right)}{4}\)
\(\Rightarrow\dfrac{12x-8y}{16}=\dfrac{6z-12x}{9}=\dfrac{8y-6z}{4}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{12x-8y}{16}=\dfrac{6z-12x}{9}=\dfrac{8y-6z}{4}=\dfrac{12x-8y+6z-12x+8y-6z}{16+9+4}=\dfrac{0}{29}=0\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{2}=\dfrac{y}{3}\\\dfrac{x}{2}=\dfrac{z}{4}\\\dfrac{y}{3}=\dfrac{z}{4}\end{matrix}\right.\Rightarrow\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}\left(đpcm\right)\)
\(\dfrac{3x-2y}{4}=\dfrac{2z-4x}{3}=\dfrac{4y-3z}{2}\\ \Rightarrow\dfrac{12x-8y}{16}=\dfrac{6z-12x}{9}=\dfrac{8y-6z}{4}\\ =\dfrac{\left(12x-8y\right)+\left(6z-12x\right)+\left(8y-6z\right)}{16+9+4}=\dfrac{0}{29}=0\\ \Rightarrow3x=2y;2z=4x;4y=3z\\ \Rightarrow\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}\)
a)vì\(\dfrac{x}{3}\)=\(\dfrac{y}{4}\)=\(\dfrac{z}{5}\)=>\(\dfrac{2x}{6}\)=\(\dfrac{3y}{12}\)=\(\dfrac{5z}{25}\)và 2x+3y+5z=86
áp dụng tính chất của dãy tỉ số bằng nhau ta có
\(\dfrac{2x}{6}\)=\(\dfrac{3y}{12}\)=\(\dfrac{5z}{25}\)=\(\dfrac{2x+3y+5z}{6+12+25}\)\(\dfrac{86}{43}\)=2
vì\(\dfrac{2x}{6}\)=2=>2x=2.6=12=>x=12:2=6
\(\dfrac{3y}{12}\)=2=>3y=12.2=24=>y=24:3=8
\(\dfrac{5z}{25}\)=2=>5z=25.2=50=>z=50:5=10
vậy x=6,y=8,z=10
vì\(\dfrac{x}{3}\)=\(\dfrac{y}{4}\)=>\(\dfrac{x}{9}\)=\(\dfrac{y}{12}\)(1)
\(\dfrac{y}{6}\)=\(\dfrac{z}{8}\)=>\(\dfrac{y}{12}\)=\(\dfrac{z}{16}\)(2)
từ (1)(2)=>\(\dfrac{x}{9}\)=\(\dfrac{y}{12}\)=\(\dfrac{z}{16}\)=>\(\dfrac{3x}{27}\)=\(\dfrac{2y}{24}\)=\(\dfrac{z}{16}\)và 3x-2y-z=13
áp dụng tính chất của dãy tỉ số bằng nhau ta có
\(\dfrac{3x}{27}\)=\(\dfrac{2y}{24}\)=\(\dfrac{z}{16}\)=\(\dfrac{3x-2y-z}{27-24-16}\)=\(\dfrac{13}{-13}\)=-1
vì\(\dfrac{3x}{27}\)=-1=>3x=-1.27=-27=>x=-27x;3=-9
\(\dfrac{2y}{24}\)=-1=>2y=-1.24=-24=>y=-24:2=-12
\(\dfrac{z}{16}\)=-1=>z=-1.16=-16
vậy...
cậu áp dụng tính chất dãy tỉ số bằng nhau đi