Question

Tìm x,y biết \(\frac{1+3x}{12}=\frac{1+6x}{16}=\frac{1-5x}{4y}\)

Answer 1

Ta có: \(\dfrac{1+3x}{12}=\dfrac{2\left(1+3x\right)}{24}=\dfrac{2+6x}{24}\)

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

\(\dfrac{2+6x}{24}=\dfrac{1+6x}{16}=\dfrac{2+6x-1-6x}{24-16}=\dfrac{1}{8}\)

Khi đó \(\dfrac{1+3x}{12}=\dfrac{1}{8}\)

\(\Rightarrow\left(1+3x\right)8=12\)

\(\Rightarrow8+24x=12\)

\(\Rightarrow24x=4\)

\(\Rightarrow x=\dfrac{1}{6}\)

Thay \(x=\dfrac{1}{6}\) vào đề bài:

\(\dfrac{1-5.\dfrac{1}{6}}{4y}=\dfrac{1}{8}\)

\(\Rightarrow\dfrac{\dfrac{1}{6}}{4y}=\dfrac{1}{8}\)

\(\Rightarrow4y=\dfrac{4}{3}\)

\(\Rightarrow y=\dfrac{1}{3}\)

Vậy \(\left[{}\begin{matrix}x=\dfrac{1}{6}\\y=\dfrac{1}{3}\end{matrix}\right.\).