\(B=\left[\dfrac{11}{4}\cdot\left[-\dfrac{5}{9}-\dfrac{4}{9}\right]\right]\cdot\dfrac{8}{33}=\dfrac{11}{4}\cdot\dfrac{8}{33}\cdot\left(-1\right)=\dfrac{-2}{3}\)

\(=\dfrac{11}{4}\cdot\left(-1\right)\cdot\dfrac{8}{33}=\dfrac{2}{3}\cdot\left(-1\right)=-\dfrac{2}{3}\)

1+2-3-4+5+6-7-8+9+................+33

=1+(2-3-4+5)+(6-7-8+9)+............+(30-31-32+33)

=1+0+0+..............+0

=1

Cảm ơn nha ! Kết bạn với tớ nhé!

1+2-3-3+5+6-7-8+9+10-11-12+13+...............+33

= 1 + ( 2-3-4+5)+(6-7-8+9)+......+(30-31-32+33)

=1

1,35326087

-2/3

0

-6/29/44

-9

nha bạn

\(A=\left(\dfrac{11}{4}.\dfrac{-5}{9}-\dfrac{4}{9}:\dfrac{4}{11}\right).\dfrac{8}{33}\)

\(=\left(\dfrac{11}{4}.\dfrac{-5}{9}-\dfrac{4}{9}.\dfrac{11}{4}\right).\dfrac{8}{33}\)

\(=\left[\dfrac{11}{4}.\left(\dfrac{-5}{9}-\dfrac{4}{9}\right)\right].\dfrac{8}{33}\)

\(=\dfrac{11}{4}.\left(-1\right).\dfrac{8}{33}\)

\(=\dfrac{11.\left(-1\right).8}{4.33}=\dfrac{11.\left(-1\right).2.4}{4.3.11}=\dfrac{-2}{3}\)

ta có \(\frac{\frac{4}{5}-\frac{4}{11}+\frac{4}{13}}{\frac{9}{5}-\frac{9}{11}+\frac{9}{13}}+\frac{\frac{5}{7}+\frac{5}{11}-\frac{5}{13}}{\frac{9}{7}+\frac{9}{11}-\frac{9}{13}}\)

       \(=\frac{4\left(\frac{1}{5}-\frac{1}{11}+\frac{1}{13}\right)}{9\left(\frac{1}{5}-\frac{1}{11}+\frac{1}{13}\right)}+\frac{5\left(\frac{1}{7}+\frac{1}{11}-\frac{1}{13}\right)}{9\left(\frac{1}{7}+\frac{1}{11}-\frac{1}{13}\right)}\)

           \(=\frac{4}{9}+\frac{5}{9}=\frac{9}{9}=1\)

coi lại câu a

1.1+1.1+1.1+1.1+1.1/1.1+1.1+9.3+6.8+.2