Theo đề bài ta có:
\(\dfrac{4}{3x-2y}=\dfrac{3}{2z-4x}=\dfrac{2}{4y-3z}\)
\(\Rightarrow\)4(2z-4x) = 3(3x-2y)
3(4y-3z) = 2(2z-4x)
Ta có:
4(2z-4x) = 3(3x-2y)\(\Rightarrow\)8z-16x = 9x-6y\(\Rightarrow y=\dfrac{25x-8z}{6}\) (1)
\(\dfrac{3}{2z-4x}=\dfrac{2}{4y-3z}\Rightarrow3\left(4y-3z\right)=2\left(2z-4x\right)\)
\(\Rightarrow12y-9z=4z-8x\Rightarrow12y+8x=13z\) (2)
Thay (1) vào (2) ta có:
2(25x-8z)+8x = 13z\(\Rightarrow\)58x = 29z\(\Rightarrow\)z = 2x\(\Rightarrow\)y = \(\dfrac{3}{2}x\)
Thay vào đề bài x + y- z= - 10 ta tìm được:
x = -10; y = -20; z = -30
Ta có : \(\frac{4}{3x-2y}=\frac{3}{2z-4x}=\frac{2}{4y-3z}\) với x+y-z = -10 (1)
\(\Rightarrow4\left(2z-4x\right)=3\left(3x-2y\right)\) ; \(3\left(4y-3z\right)=2\left(2z-4x\right)\)
Ta có :
+) \(4\left(2z-4x\right)=3\left(3x-2y\right)\Rightarrow8z-16x=9x-6y\)\(\Rightarrow y=\frac{25x-8z}{y}\left(2\right)\)
+) \(3\left(4y-3z\right)=2\left(2z-4x\right)\Rightarrow12y-9z=4z-8x\)\(\Rightarrow12y+8x=13z\left(3\right)\)
Thay (1) vào (2) ta có :
\(2\left(25x-8z\right)+8x=13z\)
\(\Rightarrow50x-16z+8x=13z\)
\(\Rightarrow58x=29z\)
\(\Rightarrow2x=z\) (4)
\(\Rightarrow y=\frac{3}{2}x\) (5)
thay (4) và (5) vào biểu thức x+y-z = -10 ta có :
\(x+y-z=-10\Leftrightarrow x+\frac{3}{2}x-2x=-10\)
\(\Rightarrow\frac{1}{2}x=-10\)
\(\Rightarrow x=-20\) ; \(y=\frac{3}{2}\left(-20\right)=-30\) ; \(z=-20\cdot2=-40\)
vậy \(x=-20;y=-30;z=-40\)
