a) ta có: \(\frac{2x}{5}=\frac{4y}{3}=\frac{3z}{10}\Rightarrow\frac{1}{12}\cdot\frac{2x}{5}=\frac{1}{12}\cdot\frac{4y}{3}=\frac{1}{12}\cdot\frac{3z}{10}\)

                                                        \(=\frac{x}{30}=\frac{y}{9}=\frac{z}{40}\)

ADTCDTSBN

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rùi bn tự lm típ nha

Ta có:

\(\frac{2x}{5}=\frac{4y}{3}=\frac{3z}{10}\)=>\(\frac{2x}{60}=\frac{4y}{36}=\frac{3z}{120}\)=>\(\frac{x}{30}=\frac{y}{9}=\frac{z}{40}\)

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

\(\frac{x}{30}=\frac{y}{9}=\frac{z}{40}=\frac{x+y+z}{30+9+40}=\frac{39,5}{79}=\frac{1}{2}\)

=>\(\frac{x}{30}=\frac{1}{2}\)=>x=15

    \(\frac{y}{9}=\frac{1}{2}\)=>y=4,5

     \(\frac{z}{40}=\frac{1}{2}\)=>z=20

Vậy x=15; y=4,5; z=20

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

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

\(=\frac{4.\left(3x-2y\right)}{4.4}=\frac{3.\left(2z-4x\right)}{3.3}=\frac{2.\left(4y-3z\right)}{2.2}\)

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

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

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

\(\Rightarrow\begin{cases}12x-8y=0\\6z-12x=0\\8y-6z=0\end{cases}\)\(\Rightarrow\begin{cases}12x=8y\\6z=12x\\8y=6z\end{cases}\)\(\Rightarrow12x=8y=6z\)

= \(\frac{x}{\frac{1}{12}}=\frac{y}{\frac{1}{8}}=\frac{z}{\frac{1}{6}}\)

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

\(\frac{x}{\frac{1}{12}}=\frac{y}{\frac{1}{8}}=\frac{z}{\frac{1}{6}}=\frac{x+y-z}{\frac{1}{12}+\frac{1}{8}-\frac{1}{6}}=\frac{-10}{\frac{1}{24}}=-10.24=-240\)

\(\Rightarrow\begin{cases}x=-240.\frac{1}{12}=-20\\y=-240.\frac{1}{8}=-30\\z=-240.\frac{1}{6}=-40\end{cases}\)

Vậy x = -20; y = -30; z = -40

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

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

=> \(\frac{4\left(3x-2y\right)}{4\cdot4}=\frac{3\left(2z-4x\right)}{3\cdot3}=\frac{2\left(4y-3z\right)}{2\cdot2}\)

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

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

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

=> \(\hept{\begin{cases}12x-8y=0\\6z-12x=0\\8y-6z=0\end{cases}}\Rightarrow\hept{\begin{cases}12x=8y\\6z=12x\\8y=6z\end{cases}}\Rightarrow12x=8y=6z\)

=> \(\frac{x}{\frac{1}{12}}=\frac{y}{\frac{1}{8}}=\frac{z}{\frac{1}{6}}\)

Đặt \(\frac{x}{\frac{1}{12}}=\frac{y}{\frac{1}{8}}=\frac{z}{\frac{1}{6}}=k\Rightarrow\hept{\begin{cases}x=\frac{1}{12}k\\y=\frac{1}{8}k\\z=\frac{1}{6}k\end{cases}}\Rightarrow\hept{\begin{cases}x=\frac{k}{12}\\y=\frac{k}{8}\\z=\frac{k}{6}\end{cases}}\)

=> \(x+y-z=\frac{k}{12}+\frac{k}{8}-\frac{k}{6}\)

=> \(\frac{k}{24}=-10\)

=> \(k=-240\)

Từ đó suy ra : \(x=-\frac{240}{12}=-20\),y = \(-\frac{240}{8}=-30\),z = \(-\frac{240}{6}=-40\)

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

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

\(=\frac{4\left(3x-2y\right)}{4.4}=\frac{3\left(2z-4x\right)}{3.3}=\frac{2\left(4y-3z\right)}{2.2}\)

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

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

\(\Rightarrow3x-2y=0\)    (1)

\(2z-4x=0\)

\(4y-3z=0\)       (2)

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

\(\Rightarrow\frac{x}{2}=-10\Rightarrow x=-20\)

\(\frac{y}{3}=-10\Rightarrow y=-30\)

\(\frac{z}{4}=-10\Rightarrow z=-40\)

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