2A=1-1/2+1/2^2-...+1/2^98-1/2^99

=>3A=1-1/2^100

=>\(A=\dfrac{2^{100}-1}{3\cdot2^{100}}\)

\(U_n=\dfrac{\left(n^2-1\right)}{n\left(n+2\right)}U_{n-1}\Rightarrow n\left(n+2\right).U_n=\left(n-1\right)\left(n+1\right).U_{n-1}\)

Đặt \(n\left(n+2\right).U_n=V_n\Rightarrow V_{n-1}=\left(n-1\right)\left(n+2-1\right).U_{n-1}=\left(n-1\right).\left(n+1\right)U_{n-1}\)

\(\Rightarrow V_n=V_{n-1}\)

\(\Rightarrow V_n=V_{n-1}=V_{n-2}=...=V_1\)

Có \(V_1=1.\left(1+2\right).U_1=1\)

\(\Rightarrow V_n=1\)

\(\Rightarrow U_n=\dfrac{V_n}{n\left(n+2\right)}=\dfrac{1}{n\left(n+2\right)}\)

\(\Rightarrow A=\dfrac{1}{1.3}+\dfrac{1}{2.4}+\dfrac{1}{3.5}+...+\dfrac{1}{2015.2017}\)

\(=\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{2015}-\dfrac{1}{2017}\right)\)

\(=\dfrac{1}{2}\left(1+\dfrac{1}{2}-\dfrac{1}{2016}-\dfrac{1}{2017}\right)\)

\(=...\)

\(E=\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{8}+\dfrac{1}{2}+\dfrac{1}{12}\)

\(E=\left(\dfrac{1}{2}+\dfrac{1}{2}\right)+\left(\dfrac{1}{3}+\dfrac{1}{6}\right)+\left(\dfrac{1}{8}+\dfrac{1}{12}+\dfrac{1}{24}\right)\)

\(E=\dfrac{2}{2}+\dfrac{3}{6}+\left(\dfrac{1}{8}+\dfrac{3}{24}\right)\)

\(E=1+\dfrac{1}{2}+\left(\dfrac{1}{8}+\dfrac{1}{8}\right)\)

\(E=\left(\dfrac{2}{2}+\dfrac{1}{2}\right)+\dfrac{2}{8}\)

\(E=\dfrac{3}{2}+\dfrac{1}{4}\)

\(E=\dfrac{6}{4}+\dfrac{1}{4}\)

\(E=\dfrac{7}{4}\)

\(=\dfrac{3}{4}-\dfrac{5}{6}\times\dfrac{7}{24}\times\dfrac{12}{7}=\dfrac{3}{4}-\dfrac{5}{12}=\dfrac{1}{3}\)

\(\dfrac{3}{4}-\dfrac{5}{6}\left(\dfrac{1}{6}+\dfrac{1}{8}\right):\dfrac{7}{12}\)

\(=\dfrac{3}{4}-\dfrac{5}{6}\cdot\dfrac{7}{24}\cdot\dfrac{12}{7}\)

\(=\dfrac{3}{4}-\dfrac{5}{12}\)

\(=\dfrac{4}{12}=\dfrac{1}{3}\)

1/3

\(A=\dfrac{6}{7}+\dfrac{1}{7}.\dfrac{2}{7}+\dfrac{1}{7}.\dfrac{5}{7}.\)

\(A=\dfrac{6}{7}+\dfrac{1}{7}\left(\dfrac{2}{7}+\dfrac{5}{7}\right).\)

\(A=\dfrac{6}{7}+\dfrac{1}{7}.1.\)

\(A=\dfrac{6}{7}+\dfrac{1}{7}=1.\)

Vậy \(A=1.\)

\(B=\dfrac{40}{9}.\dfrac{13}{3}-\dfrac{4}{3}.\dfrac{40}{9}.\)

\(B=\dfrac{4}{9}.\dfrac{13}{3}-\dfrac{4}{9}.\dfrac{40}{3}.\)

\(B=\dfrac{4}{9}\left(\dfrac{13}{3}-\dfrac{40}{3}\right).\)

\(B=\dfrac{4}{9}.\left(-9\right).\)

\(B=-4.\)

Vậy \(B=-4.\)

\(=\dfrac{3\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{7}+\dfrac{1}{11}\right)}{13\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{17}+\dfrac{1}{11}\right)}=\dfrac{3}{13}\)

\(\dfrac{\dfrac{3}{4}-\dfrac{3}{5}+\dfrac{3}{7}+\dfrac{3}{11}}{\dfrac{13}{4}-\dfrac{13}{5}+\dfrac{13}{7}+\dfrac{13}{11}}\)

\(=\dfrac{3\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{7}+\dfrac{1}{11}\right)}{13\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{7}+\dfrac{1}{11}\right)}\)

\(=\dfrac{3}{13}\)

1.\(\dfrac{4}{5}\)

2.\(\dfrac{2}{3}\)

3.\(\dfrac{39}{55}\)

a) \(\dfrac{3}{5}\times\dfrac{2}{3}:\dfrac{1}{2}=\dfrac{3}{5}\times\dfrac{2}{3}\times2=\dfrac{6}{15}\times2=\dfrac{12}{15}=\dfrac{4}{5}\)

b) \(\dfrac{2}{3}\cdot\dfrac{1}{2}+\dfrac{1}{3}=\dfrac{1}{3}+\dfrac{1}{3}=\dfrac{2}{3}\)

c) \(\left(\dfrac{5}{7}+\dfrac{2}{5}\right):\dfrac{11}{7}=\left(\dfrac{25}{35}+\dfrac{14}{35}\right)\cdot\dfrac{7}{11}=\dfrac{39}{35}\cdot\dfrac{7}{11}=\dfrac{273}{385}=\dfrac{39}{55}\)

a) = \(\dfrac{6}{15}\) \(\div\dfrac{1}{2}=\) \(\dfrac{12}{15}\) rút gọn = \(\dfrac{4}{5}\)

b) = \(\dfrac{2}{6}+\dfrac{1}{3}\) = \(\dfrac{6}{18}+\dfrac{6}{18}\) = \(\dfrac{12}{18}\) = \(\dfrac{2}{3}\)

c) = \(\dfrac{5}{7}+\dfrac{2}{5}=\dfrac{25}{35}+\dfrac{14}{35}=\dfrac{39}{35}\) \(\div\dfrac{11}{7}=\dfrac{39}{35}\)

\(=\dfrac{2}{3}+\dfrac{1}{3}.\dfrac{7}{18}.\dfrac{12}{7}\)

\(=\dfrac{2}{3}+\dfrac{7.3.2.2}{3.7.3.2.3}\)

\(=\dfrac{2}{3}+\dfrac{2}{9}=\dfrac{8}{9}\)

TICK CHO MÌNH NHÉ

Giải:

\(\dfrac{2}{3}\) + \(\dfrac{1}{3}\) . (\(-\dfrac{4}{9}\) + \(\dfrac{5}{6}\) ) : \(\dfrac{7}{12}\)

 = \(\dfrac{2}{3}\)  + \(^{\dfrac{1}{3}}\) . \(\dfrac{7}{18}\) : \(\dfrac{7}{12}\)

 = \(\dfrac{2}{3}\) + \(\dfrac{7}{54}\) : \(\dfrac{7}{12}\)

 = \(\dfrac{2}{3}\) + \(\dfrac{2}{9}\)

 = \(\dfrac{8}{9}\)

2 /3 + 1/ 3 . ( − 4 9 + 5 6 ) : 7 /12

= 2/ 3 + 1 /3 . 7 /18 . 12/ 7

= 2/ 3 + 7 /48 . 12 /7

= 2/ 3 + 1/ 4

= 11/ 12

\(=\dfrac{2}{3}+\dfrac{1}{3}.\left(\dfrac{7}{18}\right):\dfrac{7}{12}\)

\(=\dfrac{2}{3}+\dfrac{7}{54}:\dfrac{7}{12}\)

\(=\dfrac{2}{3}+\dfrac{2}{9}\)

\(=\dfrac{8}{9}\)