cho 2(x+y)=5(y+x)=3(z+x).Chứng minh x-y/4=y-z/5

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1. Ta có:

   $2(x+y) = 5(y+z) = 3(z+x)$

   $+) \quad 2(x+y) = 3(z+x)$

   $\to \dfrac{x+y}{3} = \dfrac{z+x}{2}$

   $\to \dfrac{x+y}{3} = \dfrac{z+x}{2} = \dfrac{x +y – z -x}{3 – 2} = \dfrac{y-z}{1} = y -z$

   $\to \dfrac{z+x}{2} = y – z$

   $\to \dfrac{z+x}{10} = \dfrac{y-z}{5}\qquad (1)$

   $+) \quad 5(y+z) = 3(z+x)$

   $\to \dfrac{y+z}{3} = \dfrac{z+x}{5}$

   $\to \dfrac{y+z}{3} = \dfrac{z+x}{5} = \dfrac{z+x- y- z}{5 – 3} = \dfrac{x-y}{2}$

   $\to \dfrac{z+x}{5} = \dfrac{x-y}{2}$

   $\to \dfrac{z+x}{10} = \dfrac{x-y}{4}\qquad (2)$

   $(1)(2)\to \dfrac{x-y}{4} = \dfrac{y-z}{5}$

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2. Bổ xung cách 2

    Có `2(x+y)=5(y+z)=3(z+x)`

   `⇒\frac{x+y}{15}=\frac{y+z}{6}=\frac{z+x}{10}`

   Áp dụng T/C dãy tỉ số = nhau

   `\frac{x+y}{15}=\frac{y+z}{6}=\frac{z+x}{10}=\frac{x+y-z+x}{15-10}=\frac{y-z}{5}`

   Do đó:`\frac{z+x}{10}=\frac{y-z}{5}(1)`

   Xét `\frac{x+y}{15}=\frac{y+z}{6}=\frac{z+x}{10}`

   Áp dụng T/C dãy tỉ số = nhau

   `\frac{x+y}{15}=\frac{y+z}{6}=\frac{z+x}{10}=\frac{z+x-y-z}{10-6}=\frac{x-y}{4}`

   Do đó:`\frac{z+x}{10}=\frac{x-y}{4}(2)`

   Từ `(1)(2)⇒\frac{x-y}{4}=\frac{y-z}{5}`

   Vậy đpcm

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