a) \(\dfrac{x^3}{8}=\dfrac{y^3}{64}=\dfrac{z^3}{216}\)

Từ \(\dfrac{x^3}{8}=\dfrac{y^3}{64}=\dfrac{z^3}{216}\Rightarrow\dfrac{x^3}{2^3}=\dfrac{y^3}{4^3}=\dfrac{z^3}{6^3}\)

\(\Leftrightarrow\dfrac{x^2}{2^2}=\dfrac{y^2}{4^2}=\dfrac{z^2}{6^2}\Leftrightarrow\dfrac{x^2}{4}=\dfrac{y^2}{16}=\dfrac{z^2}{36}\)

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

\(\dfrac{x^2}{4}=\dfrac{y^2}{16}=\dfrac{z^2}{36}=\dfrac{x^2+y^2+z^2}{4+16+36}=\dfrac{14}{56}=\dfrac{1}{4}\)

\(\Rightarrow\dfrac{x^2}{4}=\dfrac{1}{4}\Rightarrow x^2=\dfrac{1}{4}\cdot4\Rightarrow x^2=1\Rightarrow x=1\)

\(\dfrac{y^2}{16}=\dfrac{1}{4}\Rightarrow y^2=\dfrac{1}{4}\cdot16\Rightarrow y^2=4\Rightarrow y=2\)

\(\dfrac{z^2}{36}=\dfrac{1}{4}\Rightarrow z^2=\dfrac{1}{4}\cdot36\Rightarrow z^2=9\Rightarrow z^2=3\)

Xin lỗi mình chỉ làm được câu a)

buồn nhỉ

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

\(\Rightarrow\dfrac{4\left(3x-2y\right)}{16}=\dfrac{3\left(2z-4x\right)}{9}=\dfrac{2\left(4y-3z\right)}{4}\)

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

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

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

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

\(\dfrac{3x-2y}{4}=\dfrac{2z-4x}{3}=\dfrac{4y-3z}{2}\\ \Rightarrow\dfrac{12x-8y}{16}=\dfrac{6z-12x}{9}=\dfrac{8y-6z}{4}\\ =\dfrac{\left(12x-8y\right)+\left(6z-12x\right)+\left(8y-6z\right)}{16+9+4}=\dfrac{0}{29}=0\\ \Rightarrow3x=2y;2z=4x;4y=3z\\ \Rightarrow\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}\)

\(\dfrac{6}{11}x=\dfrac{9}{2}y=\dfrac{18}{5}z\)

\(\Leftrightarrow\dfrac{x}{\dfrac{11}{6}}=\dfrac{y}{\dfrac{2}{9}}=\dfrac{z}{\dfrac{5}{18}}\)

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

\(\dfrac{x}{\dfrac{11}{6}}=\dfrac{y}{\dfrac{2}{9}}=\dfrac{z}{\dfrac{5}{18}}=\dfrac{x+y+z}{\dfrac{11}{6}+\dfrac{2}{9}+\dfrac{5}{18}}=\dfrac{-196}{\dfrac{42}{18}}=\dfrac{-98}{\dfrac{21}{18}}=\dfrac{-588}{7}\)

(thấy lẻ,nếu đề ko sai thì làm tiếp)

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

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

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

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

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

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

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

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

\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}=\dfrac{x+y-z}{2+3-4}=\dfrac{-10}{1}=-10\)

\(\Rightarrow\left\{{}\begin{matrix}x=-10.2=-20\\y=-10.3=-30\\z=-10.4=-40\end{matrix}\right.\)

Vậy......

tks nha bn

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

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

Áp dụng t,c dãy tỉ số bằng nhau ta có :

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

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{12x-8y}{16}=0\\\dfrac{6z-12x}{9}=0\\\dfrac{8y-6z}{4}=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}12x-8y=0\\6z-12x=0\\8y-6z=0\end{matrix}\right.\)

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

Ta có

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

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

Áp dụng tính chất DTS bằng nhau

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

\(\left\{{}\begin{matrix}12x-8y=0\\6z-12x=0\\8y-6z=0\end{matrix}\right.\)

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

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

Ta có:

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

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

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

\(\Rightarrow\left\{{}\begin{matrix}12x-8y=0\\6z-12x=0\\8y-6z=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x}{2}=\dfrac{y}{3}\\\dfrac{y}{3}=\dfrac{z}{4}\\\dfrac{z}{4}=\dfrac{x}{2}\end{matrix}\right.\)

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

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

Theo đề ta có:

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

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

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

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

\(\Rightarrow\left\{{}\begin{matrix}12x=8y\\6z=12x\\8y=6z\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}12x=8y=6z\\\end{matrix}\right.\)

=> \(\dfrac{12x}{24}=\dfrac{8y}{24}=\dfrac{6z}{24}\)( MSC: 24)

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

\(\dfrac{3x-2y}{4}=\dfrac{2z-4x}{3}=\dfrac{4y-3z}{2}\\ \Rightarrow\dfrac{12x-8y}{16}=\dfrac{6z-12x}{9}=\dfrac{8y-6z}{4}\\ =\dfrac{12x-8x+6x-12x+8y-6z}{16+9+4}\\ =0\\ \Rightarrow3x=2y;2z=4x;4y=3z\\ \Rightarrow\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}\)

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

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

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

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

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

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{3x-2y}{4}=0\\\dfrac{2z-4x}{3}=0\\\dfrac{4y-3z}{2}=0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}3x=2y\\2z=4x\\4y=3z\end{matrix}\right.\)\(\Rightarrow\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}\)

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

\(\Rightarrow\dfrac{4\left(3x-2y\right)}{3.4}=\dfrac{3\left(2z-4x\right)}{3.3}=\dfrac{2\left(4y-3z\right)}{2.2}\)

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

Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{12x-8y}{12}=\dfrac{6z-12x}{9}=\dfrac{8y-6z}{4}\)

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

\(=0\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{3x-2y}{4}=0\Rightarrow3x=2y\\\dfrac{2z-4x}{3}=0\Rightarrow2z=4x\\\dfrac{4y-3z}{2}=0\Rightarrow4y=3z\end{matrix}\right.\)

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

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

a) \(\dfrac{x}{5}=\dfrac{y}{6};\dfrac{y}{8}=\dfrac{z}{7}\)và \(x+y-z=69\)

Theo đề bài, ta có:

\(\dfrac{x}{5}=\dfrac{y}{6}\Rightarrow\dfrac{x}{5}\times\dfrac{1}{8}=\dfrac{y}{6}\times\dfrac{1}{8}\Rightarrow\dfrac{x}{40}=\dfrac{y}{48}\)(1)

\(\dfrac{y}{8}=\dfrac{z}{7}\Rightarrow\dfrac{y}{8}\times\dfrac{1}{6}=\dfrac{z}{7}\times\dfrac{1}{6}\Rightarrow\dfrac{y}{48}=\dfrac{z}{42}\)(2)

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

\(\Rightarrow\dfrac{x}{40}=\dfrac{y}{48}=\dfrac{z}{42}=\dfrac{x+y-z}{40+48-42}=\dfrac{69}{46}=\dfrac{3}{2}\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{40}=\dfrac{3}{2}\Rightarrow x=\dfrac{40\times3}{2}=60\\\dfrac{y}{48}=\dfrac{3}{2}\Rightarrow y=\dfrac{48\times3}{2}=72\\\dfrac{z}{42}=\dfrac{3}{2}\Rightarrow z=\dfrac{42\times3}{2}=63\end{matrix}\right.\)

Vậy \(\Rightarrow\left\{{}\begin{matrix}x=60\\y=72\\z=63\end{matrix}\right.\)

Ta có:\(\dfrac{x}{5}=\dfrac{y}{6}\Rightarrow\dfrac{x}{20}=\dfrac{y}{24}\)(Nhân 2 vế với \(\dfrac{1}{4}\))

\(\dfrac{y}{8}=\dfrac{x}{7}\Rightarrow\dfrac{y}{24}=\dfrac{z}{21}\)(Nhân 2 vế với \(\dfrac{1}{3}\))

\(\Rightarrow\dfrac{x}{20}=\dfrac{y}{24}=\dfrac{z}{21}\)và x+y-z=6

Áp dụng tính chất dãy tỉ số bằng nhau. Ta có:

\(\dfrac{x}{20}=\dfrac{y}{24}=\dfrac{z}{21}=\dfrac{x+y-z}{20+24-21}=\dfrac{69}{23}=3\)

Vì \(\dfrac{x}{20}=3\Rightarrow x=20.3=60\)

\(\dfrac{y}{24}=3\Rightarrow y=24.3=72\)

\(\dfrac{z}{21}=3\Rightarrow z=3.21=63\)

Vậy x=60; y=72; z=63

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

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

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

\(\dfrac{12x-8y}{16}=\dfrac{6z-12x}{9}=\dfrac{8y-6z}{4}=\dfrac{12x-8y+6z-12x+8y-6z}{16+9+4}=\dfrac{0}{16+9+4}=0\)\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{2}=\dfrac{y}{3}\\\dfrac{z}{4}=\dfrac{x}{2}\\\dfrac{y}{3}=\dfrac{z}{4}\end{matrix}\right.\Rightarrow\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}\left(đpcm\right)\)