Giải hệ phương trình : Công Thức \(\left[ \begin{array}{l}\frac{1}{x} + \frac{1}{y}= \frac{1}{24}\\\frac{51}{5x} + \frac{6}{5y}= \frac{17}{8}\end{array} \right.\)
1 câu trả lời
Đáp án:
$\left\{\begin{array}{l} x=\dfrac{360}{83} \\ y=-\dfrac{90}{17}\end{array} \right..$
Giải thích các bước giải:
$\left\{\begin{array}{l} \dfrac{1}{x}+\dfrac{1}{y}=\dfrac{1}{24} \\ \dfrac{51}{5x}+\dfrac{6}{5y}=\dfrac{17}{8}\end{array} \right. (x,y \ne 0)(1)$
Đặt $\dfrac{1}{x}=u;\dfrac{1}{y}=v$
$(1) \Leftrightarrow \left\{\begin{array}{l} u+v=\dfrac{1}{24} \\ \dfrac{51}{5}u+\dfrac{6}{5}v=\dfrac{17}{8}\end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l} u=\dfrac{1}{24}-v \\ \dfrac{51}{5}\left(\dfrac{1}{24}-v\right)+\dfrac{6}{5}v=\dfrac{17}{8}\end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l} u=\dfrac{1}{24}-v \\ \dfrac{17}{40}-\dfrac{51}{5}v+\dfrac{6}{5}v=\dfrac{17}{8}\end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l} u=\dfrac{1}{24}-v \\ \dfrac{17}{40}-9v=\dfrac{17}{8}\end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l} u=\dfrac{1}{24}-v \\ -9v=\dfrac{17}{10}\end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l} u=\dfrac{1}{24}-v \\ v=-\dfrac{17}{90}\end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l} u=\dfrac{83}{360} \\ v=-\dfrac{17}{90}\end{array} \right.$
Trở lại cách đặt ta được:
$ \left\{\begin{array}{l} \dfrac{1}{x}=\dfrac{83}{360} \\ \dfrac{1}{y}=-\dfrac{17}{90}\end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l} x=\dfrac{360}{83} \\ y=-\dfrac{90}{17}\end{array} \right..$