Question

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

Answer 1

=> \(\frac{\text{2(x+y)}}{30}\)=\(\frac{\text{5(y+z)}}{30}\)=\(\frac{\text{3(z+x)}}{30}\)

=> \(\frac{\text{x+y}}{15}\)=\(\frac{\text{y+z}}{6}\)=\(\frac{\text{z+x}}{10}\)

Theo t/c dãy tỉ số bằng nhau có:

\(\frac{\text{x+y}}{15}\)=\(\frac{\text{y+z}}{6}\)=\(\frac{\text{z+x}}{10}\)=\(\frac{\left(z+x\right)-\left(y+z\right)}{10-6}\)=\(\frac{x-y}{4}\)*

\(\frac{\text{x+y}}{15}\)=\(\frac{\text{y+z}}{6}\)=\(\frac{\text{z+x}}{10}\)=\(\frac{\left(x+y\right)-\left(z+x\right)}{15-10}\)=\(\frac{y-z}{5}\)**

Từ * và ** => \(\frac{x-y}{4}\)=\(\frac{y-z}{5}\)(đpcm)

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