\(\dfrac{2x-4y}{3}=\dfrac{4z-3x}{2}=\dfrac{3y-2z}{4}\\ \Rightarrow\dfrac{6x-12y}{9}=\dfrac{8z-6x}{4}=\dfrac{12y-8z}{16}\\ =\dfrac{\left(6x-12y\right)+\left(8z-6x\right)+\left(12y-8z\right)}{9+4+16}=0\\ \Rightarrow2x=4y;4z=3x;3y=2z\\ \Rightarrow\dfrac{x}{4}=\dfrac{y}{2}=\dfrac{z}{3}=\dfrac{2x-y+z}{8-2+3}=\dfrac{27}{9}=3\\ \Rightarrow x=12;y=6;z=9\)

a, Bạn nghi sai đề: bài này mình làm rồi nên biết chỗ sai, nếu bạn nghi đúng đề thì mình làm sau nhé

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

=\(\dfrac{6x-12y}{9}=\dfrac{8z-6x}{4}=\dfrac{12y-8z}{16}\)

=\(\dfrac{6x-12y+8z-6x+12y-8z}{9+4+6}=\dfrac{0}{29}\)

Như vậy ta có thể suy ra

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

\(\Rightarrow\dfrac{x}{4}=\dfrac{y}{2}=\dfrac{z}{3}\)=\(\dfrac{2x-y+z}{8-2+3}=\dfrac{27}{9}=3\)

\(\dfrac{x}{4}=3\Rightarrow x=12\)

\(\dfrac{y}{2}=3\Rightarrow y=6\)

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

~~~~~~~~~~~~~~~~~~~~~~~~

b,

5x=8y=3z và x-2y+z= 34

Áp dụng tính chất của dãy tỉ số bằng nhau:

Ta có: 5x= 8y= 3z= \(\dfrac{5x}{120}=\dfrac{8y}{120}=\dfrac{3z}{120}=\dfrac{x}{24}=\dfrac{y}{15}=\dfrac{z}{40}\)

\(\Rightarrow\dfrac{x}{24}=\dfrac{y}{15}=\dfrac{z}{40}=\dfrac{x-2y+z}{24-30+40}=\dfrac{34}{34}=1\)

\(\dfrac{x}{24}=1\Rightarrow x=24\)

\(\dfrac{y}{15}=1\Rightarrow y=15\)

\(\dfrac{z}{40}=1\Rightarrow z=40\)

Còn 1 cách nữa nhưng thôi nha bạn

Đáp án câu b: x=24, y=15, z=40. Để xem lời giải, truy cập https://giaibaitapvenha.blogspot.com/

\(\dfrac{2x-4y}{3}=\dfrac{4z-3x}{2}=\dfrac{3y-2z}{4}\\ \Rightarrow\dfrac{6x-12y}{9}=\dfrac{8z-6x}{4}=\dfrac{12y-8z}{16}\\ =\dfrac{\left(6x-12y\right)+\left(8z-6x\right)+\left(12y-8z\right)}{4+9+16}=\dfrac{0}{29}=0\\ \Rightarrow2x=4y;4z=3x;3y=2z\\ \Rightarrow\dfrac{x}{4}=\dfrac{y}{2}=\dfrac{z}{3}\\ =\dfrac{2x-y+z}{8-2+3}=\dfrac{27}{9}=3\\ \Rightarrow x=12;y=6;z=9\)

Ta có

2x−4y3=4z−3x2=3y−2z4

⇒3(2x−4y)3.3=2(4z−3x)2.2=4(3y−2z)4.4

⇒6x−12y32=8z−6x22=12y−8z42

=6x−12y+8z−6x+12y−8z32+22+42=0

Nên 2x−4y3=0⇒2x=4y⇒x4=y2(1)

Và4z−3x2=0⇒4z=3x⇒x4=z3(2)

Từ (1) và (2) suy ra x4=y2=z3⇒2x8=y2=z3=2x+z−y8+3−2=369=4

*x4=4⇒x=4.4=16

*y2=4⇒y=2.4=8

*z3=4⇒z=3.4=12

Vậy x = 16 và y = 8 và z = 12

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

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

a) ta có : \(\dfrac{x}{2}\) = \(\dfrac{y}{3}\) = \(\dfrac{x}{16}=\dfrac{y}{24}\) ( 1)

\(\dfrac{y}{8}=\dfrac{z}{5}\) = \(\dfrac{y}{24}=\dfrac{z}{15}\) (2)

từ (1) và (2) , ta có : \(\dfrac{x}{16}=\dfrac{y}{24}=\dfrac{z}{15}\)

mà x - y + z = 35

theo tính chất của dãy tỉ số bằng nhau , ta có :

\(\dfrac{x}{16}=\dfrac{y}{24}=\dfrac{z}{15}=\dfrac{x-y+z}{16-24+15}=\dfrac{35}{7}=5\)

do đó : \(\dfrac{x}{16}=5\) => x = 5. 16 = 80

\(\dfrac{y}{24}=5\) => y = 5.24 = 120

\(\dfrac{z}{15}=5\) => z = 5.15 = 75

vậy x = 80

y = 120

z = 75

mấy câu còn lại thì tương tự nha bn

Câu b và câu e giống nhau

b)Từ 3x = 2y \(\dfrac{x}{2}=\dfrac{y}{3}\) \(\Rightarrow\dfrac{x}{10}=\dfrac{y}{15}\) (1)

Từ 5z = 2x \(\dfrac{z}{2}=\dfrac{x}{5}\) \(\Rightarrow\dfrac{x}{10}=\dfrac{z}{4}\) (2)

Từ (1) và (2) \(\Rightarrow\dfrac{x}{10}=\dfrac{y}{15}=\dfrac{z}{4}\)

Đặt \(\dfrac{x}{10}=\dfrac{y}{15}=\dfrac{z}{4}=k\)

\(\Rightarrow\left\{{}\begin{matrix}x=10k\\y=15k\\z=4k\end{matrix}\right.\)

\(\Rightarrow x\cdot y=10k\cdot15k=24\)

\(150\cdot k^2=24\)

\(k^2=\dfrac{4}{25}=\left(\pm\dfrac{2}{5}\right)^2\)

\(k=\pm\dfrac{2}{5}\)

Từ đó tìm được x ;y ;z bạn nhé

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

cứuuuuuuuuuuuu

Giải:

Ta có:

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

\(\Rightarrow\dfrac{3\left(2x-4y\right)}{3^2}=\dfrac{2\left(4z-3x\right)}{2^2}=\dfrac{4\left(3y-2z\right)}{4^2}.\)

\(\Rightarrow\dfrac{6x-12y}{9}=\dfrac{8z-6x}{4}=\dfrac{12y-8z}{16}.\)

\(=\dfrac{6x-12y+8z-6x+12y-8z}{9+4+16}.\)

\(=\dfrac{\left(6x-6x\right)+\left(8z-8z\right)+\left(12y-12y\right)}{19}=0.\)

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

\(\Rightarrow\dfrac{x}{4}=\dfrac{y}{2}=\dfrac{z}{3}\Rightarrow\dfrac{2x}{8}=\dfrac{y}{2}=\dfrac{z}{3}_{\left(1\right)}\) và \(2x-y+z=27_{\left(2\right)}.\)

Từ \(_{\left(1\right)}\) và \(_{\left(2\right)}\), kết hợp tính chất dãy tỉ số bằng nhau có:

\(\dfrac{2x}{8}=\dfrac{y}{2}=\dfrac{z}{3}=\dfrac{2x-y+z}{8-2+3}=\dfrac{27}{9}=3.\)

Từ đó: \(\left\{{}\begin{matrix}2x=3.8=24\Rightarrow x=12.\\y=3.2=6.\\z=3.3=9.\end{matrix}\right.\)

Vậy.....

\(\dfrac{2x-4y}{3}=\dfrac{4z-3x}{2}=\dfrac{3y-2z}{4}\\ \Rightarrow\dfrac{6x-12y}{9}=\dfrac{8z-6x}{4}=\dfrac{12y-8z}{16}\\ =\dfrac{6x-12y+8z-6x+12y-8z}{9+4+16}=\dfrac{0}{29}=0\\ \Rightarrow2x=4y;4z=3x;3y=2z\\ \Rightarrow\dfrac{x}{4}=\dfrac{y}{2}=\dfrac{z}{3}\\ \Rightarrow\dfrac{x}{4}=\dfrac{y}{2}=\dfrac{z}{3}=\dfrac{2x-y+z}{8-2+3}=\dfrac{27}{9}=3\\ \Rightarrow x=12;y=6;z=9\)