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24 tháng 11 2021

\(TH_1:x+y+z=0\Rightarrow\left\{{}\begin{matrix}x+y=-z\\y+z=-x\\x+z=-y\end{matrix}\right.\\ \Rightarrow Q=\dfrac{-z}{z}+\dfrac{-x}{x}+\dfrac{-y}{y}=-3\\ TH_2:x+y+z\ne0\\ \Rightarrow\dfrac{3x-2y+z}{x}=\dfrac{3y-2z+x}{y}=\dfrac{3z-2x+y}{z}=\dfrac{2x+2y+2z}{x+y+z}=2\\ \Rightarrow\left\{{}\begin{matrix}3x-2y+z=x\\3y-2z+x=y\\3z-2x+y=z\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}2x-2y=-z\\2y-2z=-x\\2z-2x=-y\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}x-y=-\dfrac{z}{2}\\y-z=-\dfrac{x}{2}\\z-x=-\dfrac{y}{2}\end{matrix}\right.\)

\(\Rightarrow Q=-\dfrac{z}{2}:z-\dfrac{x}{2}:x-\dfrac{y}{2}:y=-\dfrac{1}{2}-\dfrac{1}{2}-\dfrac{1}{2}=-\dfrac{3}{2}\)

Xét \(x+y+z=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}y+z=-x\\z+x=-y\\x+y=-z\end{matrix}\right.\)

\(\Rightarrow A=\left(2-1\right)\left(2-1\right)\left(2-1\right)=1\)

Xét \(x+y+z\ne0\) thì ta có:

\(\dfrac{x}{y+z+3x}=\dfrac{y}{z+x+3y}=\dfrac{z}{x+y+3z}=\dfrac{x+y+z}{5x+5y+5z}=\dfrac{x+y+z}{5\left(x+y+z\right)}=\dfrac{1}{5}\)

\(\Rightarrow\left\{{}\begin{matrix}5x=y+z+3x\\5y=z+x+3y\\5z=x+y+3z\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}2x=y+z\\2y=z+x\\2z=x+y\end{matrix}\right.\)

\(\Rightarrow A=\left(2+2\right)\left(2+2\right)\left(2+2\right)=64\)

Vậy \(\left[{}\begin{matrix}A=1\\A=64\end{matrix}\right.\)

Nếu bị lỗi thì bạn có thể xem đây nhé:

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11 tháng 7 2017

a)Xét \(x=\dfrac{y}{2}=\dfrac{z}{3}=k\)

\(\Rightarrow\left\{{}\begin{matrix}x=k\\y=2k\\z=3k\end{matrix}\right.\) (1)

Thay (1) vào 4x - 3y + 2z = 36

\(\Rightarrow4.k-3.2k+2.3k=36\)

\(\Rightarrow4k-6k+6k=36\Rightarrow4k=36\)

\(\Rightarrow k=\dfrac{36}{4}=9\)

\(\Rightarrow\left\{{}\begin{matrix}x=4\\y=2.4=8\\z=3.4=12\end{matrix}\right.\)

Vậy...............................................................

b) Xét \(\dfrac{x}{5}=\dfrac{y}{4}=\dfrac{z}{7}=k\)

\(\Rightarrow\left\{{}\begin{matrix}x=5k\\y=4k\\z=7k\end{matrix}\right.\) (2)

Thay (2) vào 2x - 3z = 44

\(\Rightarrow2.5k-3.7k=44\)

\(\Rightarrow-11k=44\Rightarrow k=-4\)

\(\Rightarrow\left\{{}\begin{matrix}x=5.\left(-4\right)=-20\\y=4.\left(-4\right)=-16\\z=7.\left(-4\right)=-28\end{matrix}\right.\)

Vậy,................................................

c) Xét \(\dfrac{-x}{7}=\dfrac{y}{11}=\dfrac{-z}{5}=\dfrac{x}{-7}=\dfrac{z}{-5}=k\)

\(\Rightarrow\left\{{}\begin{matrix}x=-7k\\y=11k\\z=-5k\end{matrix}\right.\) (3)

Thay (3) vào -3z - 2y - x = -88

\(\Rightarrow-3.\left(-5k\right)-2.11k-\left(-7k\right)=-88\)

\(\Rightarrow15k-22k+7k=-88\Rightarrow0k=88\)

\(\Rightarrow k\in\varnothing\)

Suy ra: Không có cặp ( x; y; z) thỏa mãn

Vậy.................................................................

d) Xét \(\dfrac{y}{12}=\dfrac{x}{-5}=\dfrac{z}{11}=k\)

\(\Rightarrow\left\{{}\begin{matrix}x=-5k\\y=12k\\z=11k\end{matrix}\right.\) (4)

Thay (4) vào 5y - 2z = 114

\(\Rightarrow6.12k-2.11k=114\)

\(\Rightarrow50k=114\Rightarrow k=2,28\)

\(\Rightarrow\left\{{}\begin{matrix}x=-5.2,28=-11,4\\y=12.2,28=27,36\\z=25,08\end{matrix}\right.\)

Vậy..............................................

e) Xét \(\dfrac{x}{25}=\dfrac{y}{17}=\dfrac{z}{32}=k\)

\(\left\{{}\begin{matrix}x=25k\\y=17k\\z=32k\end{matrix}\right.\) (5)

Thay (5) vào -2z + 3y - 4x = -452

\(\Rightarrow\left(-2\right).32k+3.17k-4.25k=-452\)

\(\Rightarrow-113k=-452\Rightarrow k=4\)

\(\Rightarrow\left\{{}\begin{matrix}x=25.5=100\\y=17.4=68\\z=32.4=128\end{matrix}\right.\)

Vậy.......................................................

11 tháng 7 2017

a) Áp dụng tính chất dãy tỉ số bằng nhau, ta có:

\(x=\dfrac{y}{2}=\dfrac{z}{3}\Rightarrow\dfrac{x}{1}=\dfrac{y}{2}=\dfrac{z}{3}\\ \Rightarrow\dfrac{4x}{4}-\dfrac{3y}{6}+\dfrac{2z}{6}=\dfrac{4x-3y+2z}{4-6+6}=\dfrac{36}{4}=9\)

+) \(\dfrac{x}{1}=9\Rightarrow x=9\)

+) \(\dfrac{y}{2}=9\Rightarrow y=18\)

+) \(\dfrac{z}{3}=9\Rightarrow z=27\)

Vậy x = 9; y = 18; z = 27.

tương tự

17 tháng 9 2021

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

17 tháng 9 2021

\(1,\Rightarrow\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}\)

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:

\(\dfrac{x}{9}=\dfrac{y}{15}=\dfrac{z}{10}=\dfrac{2x}{18}=\dfrac{3y}{45}=\dfrac{4z}{40}=\dfrac{2x+3y-4z}{18+45-40}=\dfrac{34}{23}\\ \Rightarrow\left\{{}\begin{matrix}x=\dfrac{34}{23}\cdot9=\dfrac{306}{23}\\y=\dfrac{34}{23}\cdot15=\dfrac{510}{23}\\z=\dfrac{34}{23}\cdot10=\dfrac{340}{23}\end{matrix}\right.\)

12 tháng 12 2021

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

 

 

1 tháng 2 2023

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 ... 

\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}=k\)

\(\Rightarrow x=2k;y=3k;z=4k\)

sau đó bạn tự thay vào A và B r tính nhá

7 tháng 8 2017

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 tháng 1 2021

mong  mn help khocroi

1 tháng 1 2021

Áp dụng tính chất của dãy tỉ số bằng nhau:

\(\dfrac{x+3y-z}{z}=\dfrac{y+3z-x}{x}=\dfrac{z+3x-y}{y}=\dfrac{x+3y-z+y+3z-x+z+3x-y}{x+y+z}=\dfrac{3(x+y+z)-(x+y+z)}{x+y+z}=\dfrac{2(x+y+z)}{x+y+z}=2\)

\(\Rightarrow x=y=z=0\)

\(\Rightarrow \) P không xác định. (?)

9 tháng 3 2017

TH1: \(x+y+z=0\)

\(\Rightarrow x+y=-z\)

\(y+z=-x\)

\(x+z=-y\)

\(\Rightarrow M=\dfrac{\left(x+y\right)\left(y+z\right)\left(x+z\right)}{8xyz}=\dfrac{-xyz}{8xyz}=\dfrac{-1}{8}\)

TH2: \(x+y+z\ne0\)

\(\Rightarrow2x+2y-z=3\)

\(\Rightarrow2x+2y=4z\)

\(\Rightarrow x+y=2z\)

\(x+z=2y\)

\(y+z=2x\)

\(\Rightarrow M=\dfrac{2z.2y.2x}{8xyz}=1\)

Vậy: \(M=\dfrac{-1}{8}\) hoặc \(1\)

9 tháng 3 2017

Ta có \(\dfrac{2x+2y-z}{z}=\dfrac{2x+2z-y}{y}=\dfrac{2y+2z-x}{x}\)

Áp dụng tính chất dãy tỉ số bằng nhau

\(\Rightarrow\dfrac{2x+2y-z}{z}=\dfrac{2x+2z-y}{y}=\dfrac{2y+2z-x}{x}=\dfrac{3\left(x+y+z\right)}{x+y+z}=3\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{2x+2y-z}{z}=3\\\dfrac{2x+2z-y}{y}=3\\\dfrac{2y+2z-x}{x}=3\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}2x+2y-z=3z\\2x+2z-y=3y\\2y+2z-x=3x\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}2x+2y=4z\\2x+2z=4y\\2y+2z=4x\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x+y=2z\\x+z=2y\\y+z=2x\end{matrix}\right.\)

Ta có \(M=\dfrac{\left(x+y\right)\left(y+z\right)\left(x+z\right)}{8xyz}\)

\(\Rightarrow M=\dfrac{2x.2y.2z}{8xyz}=\dfrac{8xyz}{8xyz}=1\)

Vậy \(M=1\)