\(\frac{2x-4y}{3}=\frac{4z-3x}{2}=\frac{3y-2z}{4}.\)VÀ \(2x-y+z=27\)

\(\frac{2x-4y}{3}=\frac{4z-3x}{2}=\frac{3y-2z}{4}=\frac{6x-12y}{9}\)\(=\frac{8z-6x}{4}=\frac{12y-8z}{16}\)

\(=\frac{6x-12y+8z-6x+12y-8z}{9+4+16}\)\(=\frac{0}{29}=0\)

\(\Rightarrow2x=4y\Rightarrow\frac{x}{4}=\frac{y}{2}\)

\(\Rightarrow4z=3x\Rightarrow\frac{z}{3}=\frac{x}{4}\)

\(\Rightarrow\frac{x}{4}=\frac{y}{2}=\frac{z}{3}\)

ÁP DỤNG TÍNH CHẤT CỦA DÃY TỈ SỐ BẰNG NHAU TA CÓ:

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

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

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

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

      VẬY X = 12, Y = 6, Z = 9

cứuuuuuuuuuuuu

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

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

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

mình cũng k biết

Ca thi thanh hoa k bít j thì đừng nói linh tinh

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

\(\Leftrightarrow\frac{6x-12y}{3^2}=\frac{8z-6x}{2^2}=\frac{12y-8z}{4^2}\)

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

\(\frac{6x-12y}{3^2}=\frac{8z-6x}{2^2}=\frac{12y-8z}{4^2}=\frac{6x-12y+8z-6x+12y-8z}{3^2+2^2+4^2}=0\)

\(\Leftrightarrow\hept{\begin{cases}\frac{6x-12y}{3^2}=0\\\frac{8z-6x}{2^2}=0\\\frac{12y-8z}{4^2}=0\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}2x=4y\\4z=3x\\3y=2z\end{cases}}\)  \(\Leftrightarrow\frac{x}{4}=\frac{y}{2}=\frac{z}{3}\)

\(\Leftrightarrow\frac{2x}{8}=\frac{y}{2}=\frac{z}{3}\)

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

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

\(\Leftrightarrow\hept{\begin{cases}\frac{2x}{8}=3\\\frac{y}{2}=3\\\frac{z}{3}=3\end{cases}}\)  \(\Leftrightarrow\hept{\begin{cases}x=12\\y=6\\z=9\end{cases}}\)

Vậy  \(\left(x,y,z\right)=\left(12,6,9\right)\)

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.....

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