Theo t,c dãy tỉ số bằng nhau ta có :
\(\dfrac{1+3y}{12}=\dfrac{1+5y}{5x}=\dfrac{1+7y}{4x}=\dfrac{1+5y-1+7y}{5x-4x}=\dfrac{-2y}{x}\)
\(\Leftrightarrow\dfrac{1+5y}{5x}=-\dfrac{2y}{x}\)
\(\Leftrightarrow\dfrac{1+5y}{5}=-2y\)
\(\Leftrightarrow1+5y=-2.y.5\)
\(\Leftrightarrow1+5y=-10y\)
\(\Leftrightarrow5y+10y=1\)
\(\Leftrightarrow15y=1\)
\(\Leftrightarrow y=\dfrac{1}{15}\)
\(\Leftrightarrow x=2\)
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thêm đề: tìm x, y Biết :...
ĐK: x khác 0
\(\dfrac{1+3y}{12}=\dfrac{1+5y}{5x}=\dfrac{1+7y}{4x}\\ \Leftrightarrow\dfrac{5x+15xy}{60x}=\dfrac{12+60y}{60x}=\dfrac{15+105y}{60x}\)
\(\Rightarrow5x+15xy=12+60y=15+105y\)
\(\Leftrightarrow\left\{{}\begin{matrix}5x+15xy=12+60y\\12+60y=15+105y\\15+105y=5x+15xy\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=-\dfrac{1}{15}\end{matrix}\right.\)
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mình ko bt là đúng hay sai nữa 