Question

Cho \(\dfrac{2x+1}{5}=\dfrac{3y-2}{7}=\dfrac{2x+3y-1}{6x}\). Tìm x và y

Answer 1

Theo tính chất của dãy tỉ số bằng nhau ta có:

\(\dfrac{2x+1}{5}=\dfrac{3y-2}{7}=\dfrac{2x+1+3y-2}{5+7}=\dfrac{2x+3y-1}{12}\)

Do đó: \(\dfrac{2x+3y-1}{12}=\dfrac{2x+3y-1}{6x}\)

Nếu:

\(2x+3y-1=0\Rightarrow\left\{{}\begin{matrix}2x+1=0\\3y-2=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\dfrac{-1}{2}\\y=\dfrac{2}{3}\end{matrix}\right.\)

Nếu: \(2x+3y-1\ne0\Rightarrow6x=12\Rightarrow x=2\)

Khi đó ta có:

\(\dfrac{2.2+1}{5}=\dfrac{3y-2}{7}\Rightarrow\dfrac{3y-2}{7}=1\Rightarrow y=3\)

Vậy \(x=-\dfrac{1}{2};y=\dfrac{2}{3}\) hoặc \(x=2;y=3\)