X1/19300+x2/10500=2,6.10^-5. Tìm X1 và X2 biết x1 + x2 =0,3

Đáp án:

$\begin{array}{l}
\dfrac{{{x_1}}}{{19300}} + \dfrac{{{x_2}}}{{10500}} = 2,{6.10^{ – 5}}\\
 \Rightarrow \dfrac{{{x_1}}}{{193.100}} + \dfrac{{{x_2}}}{{105.100}} = \dfrac{{2,6}}{{100000}}\\
 \Rightarrow \dfrac{{{x_1}}}{{193}} + \dfrac{{{x_2}}}{{105}} = \dfrac{{2,6}}{{1000}}\\
 \Rightarrow \dfrac{{{x_1}}}{{193}} + \dfrac{{{x_2}}}{{105}} = \dfrac{{26}}{{100}}\\
 \Rightarrow \dfrac{{{x_1}}}{{193}} + \dfrac{{{x_2}}}{{105}} = \dfrac{{13}}{{50}}\\
 \Rightarrow \dfrac{{{x_1}}}{{193}} = \dfrac{{13}}{{50}} – \dfrac{{{x_2}}}{{105}}\\
 \Rightarrow {x_1} = \dfrac{{2509}}{{50}} – \dfrac{{193{x_2}}}{{105}}\\
 \Rightarrow \dfrac{{2509}}{{50}} – \dfrac{{193{x_2}}}{{105}} + {x_2} = 0,3 = \dfrac{3}{{10}}\\
 \Rightarrow \dfrac{{88}}{{105}}{x_2} = \dfrac{{1247}}{{25}}\\
 \Rightarrow {x_2} = \dfrac{{26187}}{{440}}\\
 \Rightarrow {x_1} = \dfrac{{2509}}{{50}} – \dfrac{{193{x_2}}}{{105}} = \dfrac{{ – 5211}}{{88}}\\
Vậy\,\left( {{x_1};{x_2}} \right) = \left( {\dfrac{{ – 5211}}{{88}};\dfrac{{26187}}{{440}}} \right)
\end{array}$