Question

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

CMR: \(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}\).

Answer 1

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

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

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

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

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

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

Answer 2

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

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

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

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

Answer 3

Ta có:

\(\dfrac{3x-2y}{4}=\dfrac{12x-8y}{16}\left(1\right)\)

\(\dfrac{2z-4x}{3}=\dfrac{6z-12x}{9}\left(2\right)\)

\(\dfrac{4y-3z}{2}=\dfrac{8y-6z}{4}\left(3\right)\)

Từ \(\left(1\right);\left(2\right);\left(3\right)\Rightarrow\dfrac{12x-8y}{16}=\dfrac{6z-12x}{9}=\dfrac{8y-6z}{4}\)

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

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

+) \(\dfrac{12x-8y}{16}=0\Rightarrow12x-8y=0\cdot16=0\Rightarrow12x=8y\Rightarrow\dfrac{x}{y}=\dfrac{8}{12}=\dfrac{2}{3}\Rightarrow\dfrac{x}{2}=\dfrac{y}{3}\left(4\right)\)

+) \(\dfrac{8y-6z}{4}=0\Rightarrow8y-6z=0\cdot4=0\Rightarrow8y=6z\Rightarrow\dfrac{y}{z}=\dfrac{6}{8}=\dfrac{3}{4}\Rightarrow\dfrac{y}{3}=\dfrac{z}{4}\left(5\right)\)

Từ \(\left(4\right)\left(5\right)\Rightarrow\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}\)

Vậy \(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}\left(dpcm\right)\)

Vật vã mãi mới xong!Tick cho mk nha!