Question

Tìm x, y, z biết:

\(\dfrac{5x-1}{3}=\dfrac{7y-6}{5}=\dfrac{5x+7y-7}{4x}\)

Answer 1

Giải:

Áp dụng tính chất dãy tỉ số bằng nhau ta có:

\(\dfrac{5x-1}{3}=\dfrac{7y-6}{5}=\dfrac{5x+7y-7}{8}=\dfrac{5x+7y-7}{4x}\)

+) Xét \(5x+7y-7=0\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{5x-1}{3}=0\\\dfrac{7y-6}{5}=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}5x-1=0\\7y-6=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\dfrac{1}{5}\\y=\dfrac{6}{7}\end{matrix}\right.\)

+) Xét \(5x+7y-7\ne0\)

\(\Rightarrow4x=8\Rightarrow x=2\)

Thay \(x=2\) vào \(\dfrac{5x-1}{3}=\dfrac{7y-6}{5}\)

\(\Rightarrow3=\dfrac{7y-6}{5}\)

\(\Rightarrow7y=21\Rightarrow y=3\)

Vậy nếu \(5x+7y-7=0\) thì \(x=\dfrac{1}{5};y=\dfrac{6}{7}\)

nếu \(5x+7y-7\ne0\) thì x = 2, y = 3