tìm x;y $\frac{1+3y}{12}$+$\frac{1+5y}{5x}$+$\frac{1+7y}{4x}$ 30/09/2021 Bởi Daisy tìm x;y $\frac{1+3y}{12}$+$\frac{1+5y}{5x}$+$\frac{1+7y}{4x}$
Ta có : Ta có : $\begin{array}{l} \frac{{1 + 3y}}{{12}} = \frac{{1 + 5y}}{{5x}} = \frac{{1 + 7y}}{{4x}}\\ \Rightarrow \frac{{1 + 3y}}{{12}} = \frac{{4 + 20y}}{{20x}} = \frac{{5 + 35y}}{{20x}} = \frac{{1 + 3y + 4 + 20y – 5 – 35y}}{{12 + 20x – 20x}} = – y\\ \Rightarrow \frac{{1 + 3y}}{{12}} = – y\\ \Rightarrow 1 + 3y = – 12y\\ \Rightarrow – 15y = 1\\ \Rightarrow y = \frac{{ – 1}}{{15}}\,ta\ được\\ \frac{{5 + 35.\left( {\frac{{ – 1}}{{15}}} \right)}}{{20x}} = – \left( {\frac{{ – 1}}{{15}}} \right)\\ \Rightarrow \frac{{\frac{8}{3}}}{{20x}} = \frac{1}{{15}}\\ \Rightarrow 20x = 40\\ \Rightarrow x = 2 \end{array}$ Bình luận
Đáp án: $\begin{array}{l}Ta có:\,\frac{{1 + 3y}}{{12}} = \frac{{1 + 5y}}{{5x}} = \frac{{1 + 7y}}{{4x}}\\ \Rightarrow \frac{{1 + 3y}}{{12}} = \frac{{4 + 20y}}{{20x}} = \frac{{5 + 35y}}{{20x}} = \frac{{1 + 3y + 4 + 20y – 5 – 35y}}{{12 + 20x – 20x}} = – y\\ \Rightarrow \frac{{1 + 3y}}{{12}} = – y\\ \Rightarrow 1 + 3y = – 12y\\ \Rightarrow – 15y = 1\\ \Rightarrow y = \frac{{ – 1}}{{15}}\,ta được\\\frac{{5 + 35.\left( {\frac{{ – 1}}{{15}}} \right)}}{{20x}} = – \left( {\frac{{ – 1}}{{15}}} \right)\\ \Rightarrow \frac{{\frac{8}{3}}}{{20x}} = \frac{1}{{15}}\\ \Rightarrow 20x = 40\\ \Rightarrow x = 2\end{array}$ Vậy x=2, $y = \frac{{ – 1}}{{15}}$ Bình luận
Ta có :
Ta có : $\begin{array}{l} \frac{{1 + 3y}}{{12}} = \frac{{1 + 5y}}{{5x}} = \frac{{1 + 7y}}{{4x}}\\ \Rightarrow \frac{{1 + 3y}}{{12}} = \frac{{4 + 20y}}{{20x}} = \frac{{5 + 35y}}{{20x}} = \frac{{1 + 3y + 4 + 20y – 5 – 35y}}{{12 + 20x – 20x}} = – y\\ \Rightarrow \frac{{1 + 3y}}{{12}} = – y\\ \Rightarrow 1 + 3y = – 12y\\ \Rightarrow – 15y = 1\\ \Rightarrow y = \frac{{ – 1}}{{15}}\,ta\ được\\ \frac{{5 + 35.\left( {\frac{{ – 1}}{{15}}} \right)}}{{20x}} = – \left( {\frac{{ – 1}}{{15}}} \right)\\ \Rightarrow \frac{{\frac{8}{3}}}{{20x}} = \frac{1}{{15}}\\ \Rightarrow 20x = 40\\ \Rightarrow x = 2 \end{array}$
Đáp án:
$\begin{array}{l}
Ta có:\,\frac{{1 + 3y}}{{12}} = \frac{{1 + 5y}}{{5x}} = \frac{{1 + 7y}}{{4x}}\\
\Rightarrow \frac{{1 + 3y}}{{12}} = \frac{{4 + 20y}}{{20x}} = \frac{{5 + 35y}}{{20x}} = \frac{{1 + 3y + 4 + 20y – 5 – 35y}}{{12 + 20x – 20x}} = – y\\
\Rightarrow \frac{{1 + 3y}}{{12}} = – y\\
\Rightarrow 1 + 3y = – 12y\\
\Rightarrow – 15y = 1\\
\Rightarrow y = \frac{{ – 1}}{{15}}\,ta được\\
\frac{{5 + 35.\left( {\frac{{ – 1}}{{15}}} \right)}}{{20x}} = – \left( {\frac{{ – 1}}{{15}}} \right)\\
\Rightarrow \frac{{\frac{8}{3}}}{{20x}} = \frac{1}{{15}}\\
\Rightarrow 20x = 40\\
\Rightarrow x = 2
\end{array}$
Vậy x=2, $y = \frac{{ – 1}}{{15}}$