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\(\dfrac{3x-2y}{4}=\dfrac{2z-4x}{3}=\dfrac{4y-3z}{2}\Rightarrow\dfrac{4.\left(3x-2y\right)}{4.4}=\dfrac{3.\left(2z-4x\right)}{3.3}=\dfrac{2\left(4y-3z\right)}{2.2}\)
=\(\dfrac{12x-8y}{16}=\dfrac{6z-12x}{9}=\dfrac{8y-6z}{4}=\dfrac{0}{29}\)
\(\Rightarrow\) 12x= 8y
6z=12x
8y=6z
=> 12x=8y=6z
MSC: 24
ta có: \(\dfrac{12x}{24}=\dfrac{8y}{24}=\dfrac{6z}{24}\)= \(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}\)( đpcm)
Từ giả thiết \(\dfrac{3x-2y}{4}=\dfrac{2z-4x}{3}=\dfrac{4y-3z}{4}\)
\(\Rightarrow\dfrac{3xz-2yz}{4z}=\dfrac{2yz-4xy}{3y}=\dfrac{4xy-3xz}{4x}\)
Áp dung tính chất của dãy tỉ số bằng nhau, ta có :
\(\dfrac{3xz-2yz}{4z}=\dfrac{2yz-4xy}{3y}=\dfrac{4xy-3xz}{4x}=\dfrac{3xz-2yz+2yz-4xy+4xy-3xz}{4z+3y+4x}=\dfrac{0}{4z+3y+4x}=0\)
\(\Rightarrow3xz=2yz=4xy\)
\(\Rightarrow\dfrac{3xyz}{y}=\dfrac{2xyz}{x}=\dfrac{4xyz}{z}\)
\(\Rightarrow\dfrac{3}{y}=\dfrac{2}{x}=\dfrac{4}{z}\)
\(\Rightarrow\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}\)(đpcm)
\(\dfrac{6}{11}x=\dfrac{9}{2}y=\dfrac{18}{5}z\)
\(\Leftrightarrow\dfrac{x}{\dfrac{11}{6}}=\dfrac{y}{\dfrac{2}{9}}=\dfrac{z}{\dfrac{5}{18}}\)
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......
a) Ta có: \(6x=4y=3z\Rightarrow\dfrac{6x}{12}=\dfrac{4y}{12}=\dfrac{3z}{12}\Rightarrow\dfrac{x}{2}=\dfrac{2y}{6}=\dfrac{3z}{12}\)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:
\(\dfrac{x}{2}=\dfrac{2y}{6}=\dfrac{3z}{12}=\dfrac{x+2y-3z}{2+6-12}=\dfrac{-2}{-4}=\dfrac{1}{2}.\)
Với: \(\dfrac{x}{2}=\dfrac{1}{2}\Rightarrow x=1.\)
\(\dfrac{2y}{6}=\dfrac{y}{3}=\dfrac{1}{2}\Rightarrow y=\dfrac{1}{2}.3=\dfrac{3}{2}.\)
\(\dfrac{3z}{12}=\dfrac{z}{4}=\dfrac{1}{2}\Rightarrow z=\dfrac{1}{2}.4=\dfrac{4}{2}=2.\)
Vậy: \(x=1;y=\dfrac{3}{2};z=2.\)
\(\dfrac{3x-2y}{4}=\dfrac{2z-4x}{3}=\dfrac{4y-3z}{2}\)
\(\Rightarrow\dfrac{4\left(3x-2y\right)}{16}=\dfrac{3\left(2z-4x\right)}{9}=\dfrac{2\left(4y-3z\right)}{4}\)
\(\Rightarrow\dfrac{12x-8y}{16}=\dfrac{6z-12x}{9}=\dfrac{8y-6z}{4}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{12x-8y}{16}=\dfrac{6z-12x}{9}=\dfrac{8y-6z}{4}=\dfrac{12x-8y+6z-12x+8y-6z}{16+9+4}=\dfrac{0}{29}=0\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{2}=\dfrac{y}{3}\\\dfrac{x}{2}=\dfrac{z}{4}\\\dfrac{y}{3}=\dfrac{z}{4}\end{matrix}\right.\Rightarrow\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}\left(đpcm\right)\)
\(\dfrac{3x-2y}{4}=\dfrac{2z-4x}{3}=\dfrac{4y-3z}{2}\\ \Rightarrow\dfrac{12x-8y}{16}=\dfrac{6z-12x}{9}=\dfrac{8y-6z}{4}\\ =\dfrac{\left(12x-8y\right)+\left(6z-12x\right)+\left(8y-6z\right)}{16+9+4}=\dfrac{0}{29}=0\\ \Rightarrow3x=2y;2z=4x;4y=3z\\ \Rightarrow\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}\)
\(\dfrac{x-2}{4}=\dfrac{y+1}{5}=\dfrac{z+3}{7}\)
\(\Rightarrow\dfrac{2\left(x-2\right)}{8}=\dfrac{y+1}{5}=\dfrac{2\left(z+3\right)}{14}\)
\(\Rightarrow\dfrac{2x-4}{8}=\dfrac{y+1}{5}=\dfrac{2z+6}{14}\)
Dựa vào tính chất dãy tỉ số bằng nhau ta có:
\(=\dfrac{2x-4+y+1-2z-6}{8+5-14}\)
\(=\dfrac{2x+y-2z-9}{-1}\)
\(=\dfrac{7-9}{-1}=2\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x-2}{4}=2\Rightarrow x-2=8\Rightarrow x=10\\\dfrac{y+1}{5}=2\Rightarrow y+1=10\Rightarrow y=9\\\dfrac{z+3}{7}=2\Rightarrow z+3=14\Rightarrow z=11\end{matrix}\right.\)
\(\dfrac{3x-2y}{4}=\dfrac{2z-4x}{3}=\dfrac{4y-3z}{2}\)
\(\Leftrightarrow\dfrac{12x-8y}{16}=\dfrac{6z-12x}{9}=\dfrac{8y-6z}{4}\)
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)
\(\dfrac{3x-2y}{4}=\dfrac{2z-4x}{3}=\dfrac{4y-3z}{2}\\ \Rightarrow\dfrac{12x-8y}{16}=\dfrac{6z-12x}{9}=\dfrac{8y-6z}{4}\\ =\dfrac{12x-8x+6x-12x+8y-6z}{16+9+4}\\ =0\\ \Rightarrow3x=2y;2z=4x;4y=3z\\ \Rightarrow\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}\)
Câu 2:
\(\dfrac{x+2000}{x-2000}=\dfrac{y+2001}{y-2001}\)
\(\Leftrightarrow\left(x+2000\right)\left(y-2001\right)=\left(x-2000\right)\left(y+2001\right)\)
\(\Leftrightarrow xy-2001x+2000y-4002000=xy+2001x-2000y-4002000\)
=>-2001x+2000y=2001x-2000y
=>-4002x=-4000y
=>2001x=2000y
hay x/y=2000/2001