Cho 3x-2y : 4 = 2z – 4x : 3 = 4y – 3z : 2 . Chứng minh x : 2 = y : 3 = z : 4

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1. Đáp án:

   `3x-2y : 4 = 2z-4x : 3 = 4y-3z :2`

   `=> (3x-2y)/4=(2z-4x)/3=(4y-3z)/2`

   `=(12x-8y)/16=(6z-12x)/9=(8y-6z)/4`

   `=(12x-8y+6z-12x+8y-6z)/(12+9+4)`

   `=0`

   `=>` $\left\{\begin{matrix}3x-2y=0& \\2z-4x=0& \end{matrix}\right.$

   `=>`  $\left\{\begin{matrix}3x=2y& \\2z=4x& \end{matrix}\right.$

   `=>`  $\left\{\begin{matrix}\dfrac{x}{2}=\dfrac{y}{3}& \\\dfrac{x}{2}=\dfrac{z}{4}& \end{matrix}\right.$

   `=> x/2=y/3=z/4`

   `=> x:2=y:3=z:4`

   `=> đpcm`

    

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2. Ta có:

   `(3x – 2y) : 4 = (2z – 4x) : 3 = (4y – 3z) : 2`

   ⇒ $\dfrac{3x-2y}{4}=\dfrac{2z-4x}{3}=\dfrac{4y-3z}{2}$ 

   ⇒ $\dfrac{12x-8y}{16}=\dfrac{6z-12x}{9}=\dfrac{8y-6z}{4}$ 

   Áp dụng tính chất 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\Rightarrow 12x-8y = 0 \Rightarrow 12x = 8y $

   $\Rightarrow \dfrac{x}{8}=\dfrac{y}{12} \Rightarrow \dfrac{x}{2}=\dfrac{y}{3}$ (1)

   $\dfrac{6z-12x}{9} = 0 \Rightarrow 6z – 12x = 0 \Rightarrow 6z = 12x$

   $\Rightarrow \dfrac{x}{6}=\dfrac{z}{12} \Rightarrow \dfrac{x}{2}=\dfrac{z}{4}$ (2)

   Từ (1) và (2)

   ⇒ $\dfrac{x}{2} =\dfrac{y}{3}=\dfrac{z}{4}$ 

   $\Rightarrow$ `x : 2 = y : 3 = z : 4` (đpcm)

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