1+7y/8=1+6y/11=1+9y/2x Tìm x,y

Đáp án:

\(\left\{ \begin{array}{l}
y =  – \dfrac{3}{{29}}\\
x = 1
\end{array} \right.\)

Giải thích các bước giải:

\(\begin{array}{l}
DK:x \ne 0\\
\left\{ \begin{array}{l}
\dfrac{{1 + 7y}}{8} = \dfrac{{1 + 6y}}{{11}}\\
\dfrac{{1 + 6y}}{{11}} = \dfrac{{1 + 9y}}{{2x}}
\end{array} \right.\\
 \to \left\{ \begin{array}{l}
11 + 77y = 8 + 48y\\
2x + 12xy = 11 + 99y
\end{array} \right.\\
 \to \left\{ \begin{array}{l}
29y =  – 3\\
2x + 12xy = 11 + 99y
\end{array} \right.\\
 \to \left\{ \begin{array}{l}
y =  – \dfrac{3}{{29}}\\
2x + 12x.\left( { – \dfrac{3}{{29}}} \right) = 11 + 99.\left( { – \dfrac{3}{{29}}} \right)
\end{array} \right.\\
 \to \left\{ \begin{array}{l}
y =  – \dfrac{3}{{29}}\\
2x – \dfrac{{36}}{{29}}x = \dfrac{{22}}{{29}}
\end{array} \right.\\
 \to \left\{ \begin{array}{l}
y =  – \dfrac{3}{{29}}\\
\left( {2 – \dfrac{{36}}{{29}}} \right)x = \dfrac{{22}}{{29}}
\end{array} \right.\\
 \to \left\{ \begin{array}{l}
y =  – \dfrac{3}{{29}}\\
x = 1
\end{array} \right.
\end{array}\)