Question

1 , Tính : \(\dfrac{2}{3}.\left\{{}\begin{matrix}2\\3\end{matrix}\right.-\dfrac{2}{3}.\left[\dfrac{2}{3}-\dfrac{2}{3}.\left(\dfrac{2}{3}-\dfrac{1}{2}\right)\right]\) 2 Tìm x bik : \(0,2.\left(x-\dfrac{1}{2}\right)-0,2=x-1\)

Answer 1

1.

\(\dfrac{2}{3}.\left\{\dfrac{2}{3}-\dfrac{2}{3}.\left[\dfrac{2}{3}-\dfrac{2}{3}.\left(\dfrac{2}{3}-\dfrac{1}{2}\right)\right]\right\}\)

=\(\dfrac{2}{3}.\left\{\dfrac{2}{3}-\dfrac{2}{3}.\left[\dfrac{2}{3}-\dfrac{2}{3}.\dfrac{1}{6}\right]\right\}\)

=\(\dfrac{2}{3}.\left\{\dfrac{2}{3}-\dfrac{2}{3}.\left[\dfrac{2}{3}-\dfrac{1}{9}\right]\right\}\)

=\(\dfrac{2}{3}.\left\{\dfrac{2}{3}-\dfrac{2}{3}.\dfrac{5}{9}\right\}\)

=\(\dfrac{2}{3}.\left\{\dfrac{2}{3}-\dfrac{10}{27}\right\}\)

=\(\dfrac{2}{3}.\dfrac{8}{27}\)

=...