Question

Giải hệ phương trình sau:
\(\left\{{}\begin{matrix}\dfrac{8}{3}a+\dfrac{b}{3}=0,1\\\dfrac{8}{3}.3b+\dfrac{b}{3}=0,1\end{matrix}\right.\)

Answer 1

ta có : \(\left\{{}\begin{matrix}\dfrac{8}{3}a+\dfrac{b}{3}=0,1\\\dfrac{8}{3}3b+\dfrac{b}{3}=0,1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{8}{3}a+\dfrac{b}{3}=0,1\\\dfrac{25}{3}b=0,1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}b=\dfrac{3}{250}\\\dfrac{8}{3}a+\dfrac{\dfrac{3}{250}}{3}=0,1\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}b=\dfrac{3}{250}\\a=\dfrac{9}{250}\end{matrix}\right.\) vậy \(\left(a\overset{.}{,}b\right)=\left(\dfrac{9}{250}\overset{.}{,}\dfrac{3}{250}\right)\)