1+7y/8=1+6y/11=1+9y/2x Tìm x,y Question 1+7y/8=1+6y/11=1+9y/2x Tìm x,y in progress 0 Môn Toán Ben Gia 6 years 2020-10-25T07:11:16+00:00 2020-10-25T07:11:16+00:00 1 Answers 123 views 0
Answers ( )
Đáp án:
\(\left\{ \begin{array}{l}
y = – \dfrac{3}{{29}}\\
x = 1
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
DK:x \ne 0\\
\left\{ \begin{array}{l}
\dfrac{{1 + 7y}}{8} = \dfrac{{1 + 6y}}{{11}}\\
\dfrac{{1 + 6y}}{{11}} = \dfrac{{1 + 9y}}{{2x}}
\end{array} \right.\\
\to \left\{ \begin{array}{l}
11 + 77y = 8 + 48y\\
2x + 12xy = 11 + 99y
\end{array} \right.\\
\to \left\{ \begin{array}{l}
29y = – 3\\
2x + 12xy = 11 + 99y
\end{array} \right.\\
\to \left\{ \begin{array}{l}
y = – \dfrac{3}{{29}}\\
2x + 12x.\left( { – \dfrac{3}{{29}}} \right) = 11 + 99.\left( { – \dfrac{3}{{29}}} \right)
\end{array} \right.\\
\to \left\{ \begin{array}{l}
y = – \dfrac{3}{{29}}\\
2x – \dfrac{{36}}{{29}}x = \dfrac{{22}}{{29}}
\end{array} \right.\\
\to \left\{ \begin{array}{l}
y = – \dfrac{3}{{29}}\\
\left( {2 – \dfrac{{36}}{{29}}} \right)x = \dfrac{{22}}{{29}}
\end{array} \right.\\
\to \left\{ \begin{array}{l}
y = – \dfrac{3}{{29}}\\
x = 1
\end{array} \right.
\end{array}\)