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\(2x=3y\text{⇒}\dfrac{x}{3}=\dfrac{y}{2}\text{⇒}\dfrac{x}{21}=\dfrac{y}{14}\)
\(5y=7z\text{⇒}\dfrac{y}{7}=\dfrac{z}{5}\text{⇒}\dfrac{y}{14}=\dfrac{z}{10}\)
⇒\(\dfrac{x}{21}=\dfrac{y}{14}=\dfrac{z}{10}\)⇒\(\dfrac{3x}{63}=\dfrac{7y}{98}=\dfrac{5z}{50}\)
Áp dụng tính chất dãy tỉ số bằng nhau, ta có:
\(\dfrac{3x}{63}=\dfrac{7y}{98}=\dfrac{5z}{50}=\dfrac{3x-7y+5z}{63-98+50}=\dfrac{30}{15}=2\)
⇒x=42,y=28,z=20
\(\dfrac{x}{3}=\dfrac{y}{2}\)⇒\(\dfrac{x}{15}=\dfrac{y}{10}\)
\(\dfrac{x}{5}=\dfrac{z}{7}\text{⇒}\dfrac{x}{15}=\dfrac{z}{21}\)
⇒\(\dfrac{x}{15}=\dfrac{y}{10}=\dfrac{z}{21}\)⇒\(\dfrac{x}{15}=\dfrac{2y}{20}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{x}{15}=\dfrac{2y}{20}=\dfrac{x+2y}{15+20}=\dfrac{-112}{35}=\dfrac{-16}{5}\)
⇒x=48,y=32,z=336/5
\(a,4x=5y\:\Rightarrow\frac{x}{5}=\frac{y}{4}\Rightarrow\frac{x}{15}=\frac{y}{12}\)
\(4y=6z\Rightarrow\frac{y}{6}=\frac{z}{4}\Rightarrow\frac{y}{12}=\frac{z}{8}\)
\(\Rightarrow\frac{x}{15}=\frac{y}{12}=\frac{z}{8}\)
\(\Rightarrow\frac{x}{15}=\frac{2y}{24}=\frac{3z}{24}\)
\(\Rightarrow\frac{x-2y+3z}{15-24+24}=\frac{x}{15}=\frac{y}{12}=\frac{z}{8}\)
\(\Rightarrow\frac{5}{15}=\frac{x}{15}=\frac{y}{12}=\frac{z}{8}\)
\(\Rightarrow\frac{1}{3}=\frac{x}{15}=\frac{y}{12}=\frac{z}{8}\)
\(\Rightarrow\hept{\begin{cases}x=\frac{1}{3}\cdot15=5\\y=\frac{1}{3}\cdot12=4\\z=\frac{1}{3}\cdot8=\frac{8}{3}\end{cases}}\)
\(2x=4z\Rightarrow z=\dfrac{x}{2}\)
\(2x=-3y\Rightarrow y=\dfrac{-2}{3}x\)
Thay vào \(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=3\Leftrightarrow\dfrac{1}{x}+\dfrac{1}{\dfrac{-2}{3}x}+\dfrac{1}{\dfrac{x}{2}}=3\)
\(\Leftrightarrow\dfrac{1}{x}+\dfrac{\dfrac{-3}{2}}{\dfrac{-2}{3}.\dfrac{-3}{2}.x}+\dfrac{2}{2\dfrac{x}{2}}=3\)
\(\dfrac{1}{x}+\dfrac{\dfrac{-3}{2}}{x}+\dfrac{2}{x}\)
\(\Rightarrow\dfrac{\left(1+\dfrac{-3}{2}+2\right)}{x}=3\)
\(\Rightarrow\dfrac{\dfrac{3}{2}}{x}=3\)
\(\Rightarrow x=\dfrac{1}{2}\)
\(z=\dfrac{x}{2}=\dfrac{\dfrac{1}{2}}{2}=\dfrac{1}{4}\)
\(y=\dfrac{-2}{3}x=\dfrac{-2}{3}.\dfrac{1}{4}=\dfrac{-1}{6}\)
Vậy \(\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=\dfrac{1}{4}\\z=\dfrac{-1}{6}\end{matrix}\right.\)
\(\dfrac{x-1}{2}=\dfrac{y-2}{3}=\dfrac{z-3}{4}=\dfrac{2y-4}{6}=\dfrac{3z-9}{12}\)
Áp dụng tính chất của dãy tỉ số bằng nhau ta có :
\(\dfrac{x-1}{2}=\dfrac{y-2}{3}=\dfrac{z-3}{4}=\dfrac{2y-4}{6}=\dfrac{3z-9}{12}=\dfrac{x-1-2y+4+3z-9}{2-6+12}\)
\(=\dfrac{\left(x-2y+3z\right)+\left(-1+4-9\right)}{8}=\dfrac{14-6}{8}=\dfrac{8}{8}=1\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x-1}{2}=1\\\dfrac{y-2}{3}=1\\\dfrac{z-3}{4}=1\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x-1=2\\y-2=3\\z-3=4\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=3\\y=5\\z=7\end{matrix}\right.\)
Vậy \(\left\{{}\begin{matrix}x=3\\y=5\\z=7\end{matrix}\right.\)
1) \(x:y:z=2:3:4\) ⇒ \(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}\)
Áp dụng tính chất dãy tỉ số bằng nhau, ta có:
\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}=\dfrac{x+y+z}{2+3+4}=\dfrac{18}{9}=2\)
⇒ x=4;y=6;z=8
\(1,\Rightarrow\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}\)
Áp dụng t/c dtsbn
\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}=\dfrac{x+y+z}{2+3+4}=\dfrac{18}{9}=2\\ \Rightarrow\left\{{}\begin{matrix}x=2\cdot2=4\\y=2\cdot3=6\\z=2\cdot4=8\end{matrix}\right.\)
\(2,\) Áp dụng t/c dtsbn
\(\dfrac{x}{2}=\dfrac{y}{-3}=\dfrac{z}{4}=\dfrac{4x}{8}=\dfrac{3y}{-9}=\dfrac{2z}{8}=\dfrac{4x-3y-2z}{8-\left(-9\right)-8}=\dfrac{81}{9}=9\\ \Rightarrow\left\{{}\begin{matrix}x=2\cdot9=18\\y=2\cdot\left(-3\right)=-6\\z=2\cdot4=8\end{matrix}\right.\)
\(3,4y=3z\Rightarrow\dfrac{y}{3}=\dfrac{z}{4}\Rightarrow\dfrac{y}{6}=\dfrac{z}{8};\dfrac{x}{3}=\dfrac{y}{2}\Rightarrow\dfrac{x}{9}=\dfrac{y}{6}\\ \Rightarrow\dfrac{x}{9}=\dfrac{y}{6}=\dfrac{z}{8}\)
Áp dụng t/c dtsbn
\(\dfrac{x}{9}=\dfrac{y}{6}=\dfrac{z}{8}=\dfrac{x+y+z}{9+6+8}=\dfrac{46}{23}=2\\ \Rightarrow\left\{{}\begin{matrix}x=2\cdot9=18\\y=2\cdot6=12\\z=2\cdot8=16\end{matrix}\right.\)
\(4,5x=3y\Rightarrow\dfrac{x}{3}=\dfrac{y}{5}\Rightarrow\dfrac{x}{9}=\dfrac{y}{15};\dfrac{y}{z}=\dfrac{3}{2}\Rightarrow\dfrac{y}{3}=\dfrac{z}{2}\Rightarrow\dfrac{y}{15}=\dfrac{z}{10}\\ \Rightarrow\dfrac{x}{9}=\dfrac{y}{15}=\dfrac{z}{10}\)
Áp dụng t/c dtsbn:
\(\dfrac{x}{9}=\dfrac{y}{15}=\dfrac{z}{10}=\dfrac{2x}{18}=\dfrac{3y}{45}=\dfrac{4z}{40}=\dfrac{2x+3y-4z}{18+45-40}=\dfrac{34}{23}\\ \Rightarrow\left\{{}\begin{matrix}x=\dfrac{34}{23}\cdot9=\dfrac{306}{23}\\y=\dfrac{34}{23}\cdot15=\dfrac{510}{23}\\z=\dfrac{34}{23}\cdot10=\dfrac{340}{23}\end{matrix}\right.\)
b, Ta có : \(\dfrac{x}{3}=\dfrac{y}{4};\dfrac{y}{5}=\dfrac{z}{6}\Rightarrow\dfrac{x}{15}=\dfrac{y}{20}=\dfrac{z}{24}\)
Đặt \(x=15k;y=20k;z=24k\)
Thay vào A ta được : \(A=\dfrac{30k+60k+96k}{45k+80k+120k}=\dfrac{186k}{245k}=\dfrac{186}{245}\)
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 \(\left[{}\begin{matrix}x=-4;y=-6;z=-10\\x=4;y=6;z=10\end{matrix}\right..\)
\(\dfrac{2x}{3}=\dfrac{3y}{4}=\dfrac{4z}{5}\)
\(\Leftrightarrow\dfrac{x}{\dfrac{3}{2}}=\dfrac{y}{\dfrac{4}{3}}=\dfrac{z}{\dfrac{5}{4}}\)
\(\Leftrightarrow\dfrac{x}{\dfrac{3}{2}}=\dfrac{2y}{\dfrac{8}{3}}=\dfrac{4z}{5}\)
Theo t,c dãy tỉ số bằng nhau ta có :
\(\dfrac{x}{\dfrac{3}{2}}=\dfrac{2y}{\dfrac{8}{3}}=\dfrac{4z}{5}=\dfrac{x+2y+4z}{\dfrac{3}{2}+\dfrac{8}{3}+5}=\dfrac{220}{\dfrac{55}{6}}=24\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x}{\dfrac{3}{2}}=24\\\dfrac{2y}{\dfrac{8}{3}}=24\\\dfrac{4z}{5}=24\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=36\\y=32\\z=30\end{matrix}\right.\)
Vậy ...
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
b) Ta có: 7x=10y=12z
nên \(\dfrac{x}{\dfrac{1}{7}}=\dfrac{y}{\dfrac{1}{10}}=\dfrac{z}{\dfrac{1}{12}}\)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{x}{\dfrac{1}{7}}=\dfrac{y}{\dfrac{1}{10}}=\dfrac{z}{\dfrac{1}{12}}=\dfrac{x+y+z}{\dfrac{1}{7}+\dfrac{1}{10}+\dfrac{1}{12}}=\dfrac{685}{\dfrac{137}{420}}=2100\)
Do đó:
\(\left\{{}\begin{matrix}x=2100\cdot\dfrac{1}{2}=1050\\y=2100\cdot\dfrac{1}{10}=210\\z=2100\cdot\dfrac{1}{12}=175\end{matrix}\right.\)
a) \(\dfrac{x}{y}=-\dfrac{3}{5}\) và x-2y=-52
Ta có: \(\dfrac{x}{y}=-\dfrac{3}{5}\Rightarrow\dfrac{x}{-3}=\dfrac{y}{5}\)
Áp dụng tính chất dãy tỉ số bằng nhau
\(\dfrac{x}{-3}=\dfrac{y}{5}=\dfrac{x-2y}{\left(-3\right)-2\times5}=\dfrac{-52}{-13}=4\)( vì x-2y = -52)
Suy ra: \(\dfrac{x}{-3}=4\Rightarrow x=4\times\left(-3\right)=-12\)
\(\dfrac{y}{5}=4\Rightarrow y=4\times5=20\)
Vậy x= -12, y= 20
b)3x=y=6z và 2x+3y-4z = 90
Ta có 3x=y=6z \(\Rightarrow\dfrac{x}{2}=\dfrac{y}{6}=\dfrac{z}{1}\)
Áp dụng tính chất dãy tỉ số bằng nhau
\(\dfrac{x}{2}=\dfrac{y}{6}=\dfrac{z}{1}=\dfrac{2x+3y-4z}{2\times2+3\times6-4\times1}=\dfrac{90}{18}=5\)(vì 2x+3y-4z = 90)
Suy ra: \(\dfrac{x}{2}=5\Rightarrow x=5\times2=10\)
\(\dfrac{y}{6}=5\Rightarrow y=5\times6=30\)
\(\dfrac{z}{1}=5\Rightarrow z=5\times1=5\)
Vậy x= 10, y= 30, z = 5
còn câu c)\(\dfrac{2x}{3}=\dfrac{6y}{5}=\dfrac{4z}{3}\) và x+2y-3z=99
Ta có : \(\dfrac{2x}{3}=\dfrac{6y}{5}=\dfrac{4z}{3}\)
\(\Rightarrow\dfrac{2x}{3\times12}=\dfrac{6y}{5\times12}=\dfrac{4z}{3\times12}\)
\(\Rightarrow\dfrac{x}{18}=\dfrac{y}{10}=\dfrac{z}{9}\)
Sau đó Mai áp dụng tính chất dãy tỉ số = nhau rùi lm như trên nha