Các bn ơi giúp mk vs
\(\dfrac{2x}{3}=\dfrac{3y}{4}=\dfrac{4z}{5}\) và 2x + y - 3z= 14
1, x : y : z = 2 : 3 : 4 và x + y + z = 18
2, \(\dfrac{x}{2}=\dfrac{y}{-3}=\dfrac{z}{4}\) và 4x - 3y - 2z = 81
3, \(\dfrac{x}{3}=\dfrac{y}{2};\) 4y = 3z và x + y +z = 46
4, 5x = 3y; \(\dfrac{y}{z}=\dfrac{3}{2}\) và 2x + 3y -4z =34
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:
\(\dfrac{x}{9}=\dfrac{y}{15}=\dfrac{z}{10}=\dfrac{2x}{18}=\dfrac{3y}{45}=\dfrac{4z}{40}=\dfrac{2x+3y-4z}{18+45-40}=\dfrac{34}{23}\\ \Rightarrow\left\{{}\begin{matrix}x=\dfrac{34}{23}\cdot9=\dfrac{306}{23}\\y=\dfrac{34}{23}\cdot15=\dfrac{510}{23}\\z=\dfrac{34}{23}\cdot10=\dfrac{340}{23}\end{matrix}\right.\)
TÌM X,Y,Z BIẾT:
A.\(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\) = 3 và 2x = -3y = 4z
B.\(\dfrac{x-1}{2}=\dfrac{y-2}{3}=\dfrac{z-3}{4}\)và x- 2y +3z = 14
\(2x=4z\Rightarrow z=\dfrac{x}{2}\)
\(2x=-3y\Rightarrow y=\dfrac{-2}{3}x\)
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 \(\left\{{}\begin{matrix}x=3\\y=5\\z=7\end{matrix}\right.\)
a )\(\dfrac{x}{3}\) = \(\dfrac{y}{5}\) = \(\dfrac{z}{7}\) và 3x - 2z =15
b)\(\dfrac{x}{5}\) = \(\dfrac{4}{3}\) = \(\dfrac{z}{2}\) và 2x -3y =100
c)\(\dfrac{x}{-3}\) = \(\dfrac{4}{-5}\) \(\dfrac{z}{-4}\) và 3z -2x =36
d) \(\dfrac{x}{2}\) = y = \(^{\dfrac{z}{3}}\) và 3x -2 + 4z =16
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 ...
d: x/2=y/1=z/3
mà 3x+4z=16+2=18
nên Áp dụng tính chất của DTSBN, ta được:
\(\dfrac{x}{2}=\dfrac{y}{1}=\dfrac{z}{3}=\dfrac{3x+4z}{3\cdot2+4\cdot3}=\dfrac{18}{18}=1\)
=>x=2; y=1; z=3
a: Áp dụng tính chất của DTSBN, ta được:
\(\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{z}{7}=\dfrac{3x-2z}{3\cdot3-2\cdot7}=\dfrac{15}{-5}=-3\)
=>x=-9; y=-15; z=-21
b: Áp dụng tính chất của DTSBN, ta được:
\(\dfrac{x}{5}=\dfrac{y}{3}=\dfrac{z}{2}=\dfrac{2x-3y}{2\cdot5-3\cdot3}=\dfrac{100}{1}=100\)
=>x=500; y=300; z=200
c: Áp dụng tính chất của DTSBN, ta được:
\(\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{z}{4}=\dfrac{3z-2x}{3\cdot4-2\cdot3}=\dfrac{36}{6}=6\)
=>x=18; y=30; z=24
\(\dfrac{2x}{3}=\dfrac{3y}{4}=\dfrac{4z}{5}\) và \(x+y+z=49\)
\(\dfrac{2x}{3}=\dfrac{3y}{4}=\dfrac{4z}{5}\Rightarrow\dfrac{x}{\dfrac{3}{2}}=\dfrac{y}{\dfrac{4}{3}}=\dfrac{z}{\dfrac{5}{4}}\)
Áp dụng t/c dtsbn:
\(\dfrac{x}{\dfrac{3}{2}}=\dfrac{y}{\dfrac{4}{3}}=\dfrac{z}{\dfrac{5}{4}}=\dfrac{x+y+z}{\dfrac{3}{2}+\dfrac{4}{3}+\dfrac{5}{4}}=\dfrac{49}{\dfrac{49}{12}}=12\\ \Rightarrow\left\{{}\begin{matrix}x=12\cdot\dfrac{3}{2}=18\\y=12\cdot\dfrac{4}{3}=16\\z=12\cdot\dfrac{5}{4}=15\end{matrix}\right.\)
Hãy tìm x,y,z biết:
a) \(\dfrac{x}{y}=\dfrac{-3}{5}\) và x-2y=-52
b) 3x=y=6z và 2x+3y-4z=90
c) \(\dfrac{2x}{3}=\dfrac{6y}{5}=\dfrac{4z}{3}\) và x+2y-3z=99
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
\(2x=3y;4y=5z\) và \(2x+3y-4z=56\)
\(\dfrac{x}{3}=\dfrac{y}{7};\dfrac{y}{2}=\dfrac{z}{5}\) và x + y + z = \(-10\)
(Áp dụng t/c của dãy tỉ số bằng nhau)
1/2x=3y=4z và x-y-z=35
2/\(\dfrac{x}{3}\)=\(\dfrac{y}{4}\):\(\dfrac{y}{5}\)=\(\dfrac{z}{7}\)và 2x+3y-z=186
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....
1: Ta có: 2x=3y=4z
nên \(\dfrac{x}{\dfrac{1}{2}}=\dfrac{y}{\dfrac{1}{3}}=\dfrac{z}{\dfrac{1}{4}}\)
mà x-y-z=35
nên Áp dụng tính chất của dãy tỉ số bằng nhau,ta được:
\(\dfrac{x}{\dfrac{1}{2}}=\dfrac{y}{\dfrac{1}{3}}=\dfrac{z}{\dfrac{1}{4}}=\dfrac{x-y-z}{\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{4}}=\dfrac{35}{-\dfrac{1}{12}}=-420\)
Do đó: x=-210; y=-140; z=-105
2: Ta có: \(\dfrac{x}{3}=\dfrac{y}{4}\)
nên \(\dfrac{x}{15}=\dfrac{y}{20}\left(1\right)\)
Ta có: \(\dfrac{y}{5}=\dfrac{z}{7}\)
nên \(\dfrac{y}{20}=\dfrac{z}{28}\left(2\right)\)
Từ (1) và (2) suy ra \(\dfrac{x}{15}=\dfrac{y}{20}=\dfrac{z}{28}\)
hay \(\dfrac{2x}{30}=\dfrac{3y}{60}=\dfrac{z}{28}\)
mà 2x+3y-z=186
nên Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{2x}{30}=\dfrac{3y}{60}=\dfrac{z}{28}=\dfrac{2x+3y-z}{30+60-28}=\dfrac{186}{62}=3\)
Do đó: x=45; y=60; z=84
1/ x\(\dfrac{x}{3}=\dfrac{y}{8}=\dfrac{z}{5}\text{và}2x+3y-z=50\)
2/ x : y : z = 3 : 5 ; ( - 2 ) và 5x - y + 3z = -16
3/ 2x + 3y ; 7z = 5y và 3x - 7y + 5z = 30
4/ \(\dfrac{x}{2}=\dfrac{y}{3};\dfrac{y}{4}=\dfrac{z}{5}\text{và}x-y-z=38\)
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
Tìm x và y biết:
a) \(\dfrac{2x}{3y}=\dfrac{-1}{3}\)và -2x+3y=7
b)\(\dfrac{x}{3}=\dfrac{y}{4}\)và 2x+5y=10
c) 7x=3y và x-y=16
d)\(\dfrac{x}{5}=\dfrac{y}{1}=\dfrac{z}{-2}\)và x+y-2z=160
e)\(\dfrac{x}{10}=\dfrac{y}{5};\dfrac{y}{2}=\dfrac{z}{3}\)và 2x-3y+4z=330
f) \(\dfrac{x}{-2}=\dfrac{-y}{4}=\dfrac{z}{5}\)và x-2y+3z=1200
g) \(\dfrac{x}{3}=\dfrac{y}{8}=\dfrac{z}{5}\)và 2x+3y-z=50
a,
\(\dfrac{2x}{3y}=\dfrac{-1}{3}\\ \Rightarrow\dfrac{2x}{-1}=\dfrac{3y}{3}\\ \Leftrightarrow\dfrac{-2x}{1}=\dfrac{3y}{3}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{-2x}{1}=\dfrac{3y}{3}=\dfrac{-2x+3y}{1+3}=\dfrac{7}{4}\)
\(\dfrac{-2x}{1}=\dfrac{7}{4}\Rightarrow-2x=\dfrac{7}{4}\Rightarrow x=\dfrac{7}{4}:\left(-2\right)=\dfrac{-7}{8}\\ \dfrac{3y}{3}=\dfrac{7}{4}\Rightarrow y=\dfrac{7}{4}\)
Vậy \(x=\dfrac{-7}{8};y=\dfrac{7}{4}\)
b,
\(\dfrac{x}{3}=\dfrac{y}{4}\\ \Leftrightarrow\dfrac{2x}{6}=\dfrac{5y}{20}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{2x}{6}=\dfrac{5y}{20}=\dfrac{2x+5y}{6+20}=\dfrac{10}{26}=\dfrac{5}{13}\\ \dfrac{x}{3}=\dfrac{2x}{6}=\dfrac{5}{13}\Rightarrow x=\dfrac{5}{13}\cdot3=\dfrac{15}{13}\\ \dfrac{y}{4}=\dfrac{5y}{20}=\dfrac{5}{13}\Rightarrow y=\dfrac{5}{13}\cdot4=\dfrac{20}{13}\)
Vậy \(x=\dfrac{15}{13};y=\dfrac{20}{13}\)
c,
\(7x=3y\\ \Rightarrow\dfrac{x}{3}=\dfrac{y}{7}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{x}{3}=\dfrac{y}{7}=\dfrac{x-y}{3-7}=\dfrac{16}{-4}=-4\\ \dfrac{x}{3}=-4\Rightarrow x=\left(-4\right)\cdot3=-12\\ \dfrac{y}{7}=-4\Rightarrow y=\left(-4\right)\cdot7=-28\)
Vậy \(x=-12;y=-28\)
d,
\(\dfrac{x}{5}=\dfrac{y}{1}=\dfrac{z}{-2}\\ \Leftrightarrow\dfrac{x}{5}=\dfrac{y}{1}=\dfrac{-2z}{4}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{x}{5}=\dfrac{y}{1}=\dfrac{-2z}{4}=\dfrac{x+y+\left(-2z\right)}{5+1+4}=\dfrac{x+y-2z}{10}=\dfrac{160}{10}=16\\ \dfrac{x}{5}=16\Rightarrow x=16\cdot5=80\\ \dfrac{y}{1}=16\Rightarrow y=16\\ \dfrac{z}{-2}=\dfrac{-2z}{4}=16\Rightarrow z=16\cdot\left(-2\right)=-32\)
Vậy \(x=80;y=16;z=-32\)
e,
\(\dfrac{x}{10}=\dfrac{y}{5}\Rightarrow\dfrac{x}{20}=\dfrac{y}{10};\dfrac{y}{2}=\dfrac{z}{3}\Rightarrow\dfrac{y}{10}=\dfrac{z}{15}\\ \Rightarrow\dfrac{x}{20}=\dfrac{y}{10}=\dfrac{z}{15}\\ \Leftrightarrow\dfrac{2x}{40}=\dfrac{3y}{30}=\dfrac{4z}{60}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{2x}{40}=\dfrac{3y}{30}=\dfrac{4z}{60}=\dfrac{2x-3y+4z}{40-30+60}=\dfrac{330}{70}=\dfrac{33}{7}\)
\(\dfrac{x}{20}=\dfrac{2x}{40}=\dfrac{33}{7}\Rightarrow x=\dfrac{33}{7}\cdot20=\dfrac{660}{7}\\ \dfrac{y}{10}=\dfrac{3y}{30}=\dfrac{33}{7}\Rightarrow y=\dfrac{33}{7}\cdot10=\dfrac{330}{7}\\ \dfrac{z}{15}=\dfrac{4z}{60}=\dfrac{33}{7}\Rightarrow z=\dfrac{33}{7}\cdot15=\dfrac{495}{7}\)
Vậy \(x=\dfrac{660}{7};y=\dfrac{330}{7};z=\dfrac{495}{7}\)
f,
\(\dfrac{x}{-2}=\dfrac{-y}{4}=\dfrac{z}{5}\\ \Leftrightarrow\dfrac{x}{-2}=\dfrac{-2y}{8}=\dfrac{3z}{15}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{x}{-2}=\dfrac{-2y}{8}=\dfrac{3z}{15}=\dfrac{x+\left(-2y\right)+3z}{\left(-2\right)+8+15}=\dfrac{x-2y+3z}{21}=\dfrac{1200}{21}=\dfrac{400}{7}\)
\(\dfrac{x}{-2}=\dfrac{400}{7}\Rightarrow x=\dfrac{400}{7}\cdot\left(-2\right)=\dfrac{-800}{7}\\ \dfrac{-y}{4}=\dfrac{-2y}{8}=\dfrac{400}{7}\Rightarrow-y=\dfrac{400}{7}\cdot4=\dfrac{1600}{7}\Rightarrow y=\dfrac{-1600}{7}\\ \dfrac{z}{5}=\dfrac{3z}{15}=\dfrac{400}{7}\Rightarrow z=\dfrac{400}{7}\cdot5=\dfrac{2000}{7}\)
Vậy \(x=\dfrac{-800}{7};y=\dfrac{-1600}{7};z=\dfrac{2000}{7}\)
g,
\(\dfrac{x}{3}=\dfrac{y}{8}=\dfrac{z}{5}\\ \Leftrightarrow\dfrac{2x}{6}=\dfrac{3y}{24}=\dfrac{z}{5}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{2x}{6}=\dfrac{3y}{24}=\dfrac{z}{5}=\dfrac{2x+3y-z}{6+24-5}=\dfrac{50}{25}=2\)
\(\dfrac{x}{3}=\dfrac{2x}{6}=2\Rightarrow x=2\cdot3=6\\ \dfrac{y}{8}=\dfrac{3y}{24}=2\Rightarrow y=2\cdot8=16\\ \dfrac{z}{5}=2\Rightarrow z=2\cdot5=10\)
Vậy \(x=6;y=16;z=10\)
Làm gấp nên k có kiểm tra, bn bấm máy tính dò lại nhé