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16 tháng 8 2017

Coi đề lại câu a

b,

\(\dfrac{x-1}{2}=\dfrac{y-2}{3}=\dfrac{z-3}{4}\\ \dfrac{x-1}{2}=\dfrac{2\left(y-2\right)}{2\cdot3}=\dfrac{3\cdot\left(z-3\right)}{3\cdot4}\\ \dfrac{x-1}{2}=\dfrac{2y-4}{6}=\dfrac{3z-9}{12}\)

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

\(\dfrac{x-1}{2}=\dfrac{2y-4}{6}=\dfrac{3z-9}{12}=\dfrac{x-1-\left(2y-4\right)+3z-9}{2-6+12}=\dfrac{x-1-2y+4+3z-9}{8}=\dfrac{\left(x-2y+3z\right)+\left(4-1-9\right)}{8}=\dfrac{14+\left(-6\right)}{8}=\dfrac{8}{8}=1\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x-1}{2}=1\Rightarrow x-1=2\Rightarrow x=3\\\dfrac{2y-4}{6}=1\Rightarrow2y-4=6\Rightarrow2y=10\Rightarrow y=5\\\dfrac{3z-9}{12}=1\Rightarrow3z-9=12\Rightarrow3z=21\Rightarrow z=7\end{matrix}\right.\)

Vậy x = 3; y = 5; z = 7

16 tháng 8 2017

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

\(\Rightarrow\dfrac{x-1}{2}=\dfrac{2y-4}{6}=\dfrac{3z-9}{12}\)

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

\(\dfrac{x-1}{2}=\dfrac{2y-4}{6}=\dfrac{3z-9}{12}\)

\(=\dfrac{x-1-2y+4+3z-9}{2-6+12}\)

\(=\dfrac{14-6}{14-6}=1\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x-1}{2}=1\\\dfrac{y-2}{3}=1\\\dfrac{z-3}{4}=1\end{matrix}\right.\)

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

28 tháng 10 2021

1) Áp dụng t/c dtsbn:

\(\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{z}{7}=\dfrac{2x}{6}=\dfrac{3y}{15}=\dfrac{2x+3y-z}{6+15-7}=\dfrac{-14}{14}=-1\)

\(\Rightarrow\left\{{}\begin{matrix}x=\left(-1\right).3=-3\\y=\left(-1\right).5=-5\\z=\left(-1\right).7=-7\end{matrix}\right.\)

2) \(\Rightarrow\dfrac{x}{8}=\dfrac{y}{12}=\dfrac{z}{15}\)

Áp dụng t/c dtsbn:

\(\dfrac{x}{8}=\dfrac{y}{12}=\dfrac{z}{15}=\dfrac{x-y-z}{8-12-15}=\dfrac{28}{-19}\)

\(\Rightarrow\left\{{}\begin{matrix}x=-\dfrac{28}{19}.8=-\dfrac{224}{19}\\y=-\dfrac{28}{19}.12=-\dfrac{336}{19}\\z=-\dfrac{28}{19}.15=-\dfrac{420}{19}\end{matrix}\right.\)

28 tháng 10 2021

a, Áp dụng t/c dtsbn:

\(\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{z}{7}=\dfrac{2x+3y-z}{3\cdot2+5\cdot3-7}=\dfrac{-14}{14}=-1\\ \Leftrightarrow\left\{{}\begin{matrix}x=-3\\y=-5\\z=-7\end{matrix}\right.\)

b, \(\dfrac{x}{2}=\dfrac{y}{3};\dfrac{y}{4}=\dfrac{z}{5}\Leftrightarrow\dfrac{x}{8}=\dfrac{y}{12}=\dfrac{z}{15}\)

Áp dụng t/c dtsbn:

\(\dfrac{x}{8}=\dfrac{y}{12}=\dfrac{z}{15}=\dfrac{x-y-z}{8-12-15}=\dfrac{28}{-19}\\ \Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{224}{19}\\y=-\dfrac{336}{19}\\z=-\dfrac{420}{19}\end{matrix}\right.\)

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:

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

18 tháng 10 2018

a) Giải

\(5x=2y=3z\)

\(\Rightarrow\dfrac{5x}{30}=\dfrac{2y}{30}=\dfrac{3z}{30}\)

\(\Rightarrow\dfrac{x}{6}=\dfrac{y}{15}=\dfrac{z}{10}\)

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)

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

\(\Rightarrow\dfrac{x-1}{2}=\dfrac{y-2}{3}=\dfrac{z-3}{4}=\dfrac{2y-4}{6}=\dfrac{3z-9}{12}\)

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 \(x=\dfrac{13}{5},\) \(y=\dfrac{22}{5}\)\(z=\dfrac{31}{5}.\)

c) Giải

Đặt \(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{5}=k\Rightarrow\left\{{}\begin{matrix}x=2k\\y=3k\\z=5k\end{matrix}\right.\)

\(x^2+2y^2-z^2=-12\)

\(\Rightarrow\left(2k\right)^2+2\left(3k\right)^2-\left(5k\right)^2=-12\)

\(\Rightarrow4.k^2+18.k^2-25.k^2=-12\)

\(\Rightarrow\left(-3\right)k^2=-12\)

\(\Rightarrow k^2=4\)

\(\Rightarrow k=\pm2\)

\(\circledast k=-2\Rightarrow\left\{{}\begin{matrix}x=-4\\y=-6\\z=-10\end{matrix}\right.\)

\(\circledast k=2\Rightarrow\left\{{}\begin{matrix}x=4\\y=6\\z=10\end{matrix}\right.\)

Vậy \(\left[{}\begin{matrix}x=-4;y=-6;z=-10\\x=4;y=6;z=10\end{matrix}\right..\)

20 tháng 10 2018

câu b bạn ko làm đc hả

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 ... 

22 tháng 10 2021

d: Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

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

Do đó: x=18; y=12; z=9

22 tháng 10 2021

a) Thay x + 3y - 2z vào biểu thức ta có:

 \(\dfrac{x - 1}{3} = \dfrac{3(y + 2)}{3 . 4} = \dfrac{2(z - 2)}{2 . 3}\) = \(​​​​\dfrac{x - 1}{3} = \dfrac{3x + 6}{12} = \dfrac{2z - 4}{6}\)

Áp dụng tính chất dãy tỉ số bằng nhua ta có:

\(​​​​\dfrac{x - 1}{3} = \dfrac{3y + 6}{12} = \dfrac{2z - 4}{6} = ​​​​\dfrac{x - 1}{3}+ \dfrac{3y + 6}{12} -\dfrac{2z - 4}{6}\) 

=\(​​​​\dfrac{x - 1 + 3y + 6 - 2z + 4}{3 + 12 -6} \) = \(​​​​\dfrac{(x + 3y - 2z) + ( -1 + 6 +4)}{3 + 12 - 6} \)

=\(​​​​\dfrac{36 + 9}{9}\) = 5

=> \(​​​​\dfrac{x - 1}{3} =\) 5 => x - 1 = 5.3 =15 => x = 5+1 = 6

=>

=>

Vậy ...

(Bạn dựa theo cách này và lm những bài tiếp nhé!)

 

 

 

 

 

23 tháng 10 2021

d: Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

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

Do đó: x=18; y=12; z=9

22 tháng 10 2018

\(2x=4z\Rightarrow z=\dfrac{x}{2}\)

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

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