Toán Cho 3x-2y : 4 = 2z – 4x : 3 = 4y – 3z : 2 . Chứng minh x : 2 = y : 3 = z : 4 23/07/2021 By Eliza Cho 3x-2y : 4 = 2z – 4x : 3 = 4y – 3z : 2 . Chứng minh x : 2 = y : 3 = z : 4
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` Trả lời
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) Trả lời
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`
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)