\(=5\left(\dfrac{5}{1\cdot6}+\dfrac{5}{6\cdot11}+...+\dfrac{5}{101\cdot106}\right)\\ =5\left(1-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{11}+...+\dfrac{1}{101}-\dfrac{1}{106}\right)\\ =5\left(1-\dfrac{1}{106}\right)=5\cdot\dfrac{105}{106}=\dfrac{525}{106}\)

cảm ơn bạn

\(B=\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+\frac{1}{3\cdot4\cdot5}+\frac{1}{18\cdot19\cdot20}\)

\(B=\frac{1}{2}\left(\frac{2}{1\cdot2\cdot3}+\frac{2}{2\cdot3\cdot4}+\frac{2}{3\cdot4\cdot5}+\frac{2}{18\cdot19\cdot20}\right)\)

\(B=\frac{1}{2}\left(\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+\frac{1}{3\cdot4}-\frac{1}{4\cdot5}+...+\frac{1}{18\cdot19}-\frac{1}{19\cdot20}\right)\)

\(B=\frac{1}{2}\left(\frac{1}{1\cdot2}-\frac{1}{19\cdot20}\right)\)

\(B=\frac{1}{2}\cdot\frac{189}{380}=\frac{189}{760}\)

\(C=\frac{52}{1\cdot6}+\frac{52}{6\cdot11}+\frac{52}{11\cdot16}+...+\frac{52}{31\cdot36}\)

\(C=\frac{52}{5}\left(\frac{5}{1\cdot6}+\frac{5}{6\cdot11}+\frac{5}{11\cdot16}+...+\frac{6}{31\cdot36}\right)\)

\(C=\frac{52}{5}\left(1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}+...+\frac{1}{31}-\frac{1}{36}\right)\)

\(C=\frac{52}{5}\cdot\left(1-\frac{1}{36}\right)\)

\(C=\frac{91}{9}\)

Ta có : 

\(A=\frac{5^2}{6.11}+\frac{5^2}{11.16}+\frac{5^2}{16.21}+...+\frac{5^2}{91.96}\)

\(A=5\left(\frac{5}{6.11}+\frac{5}{11.16}+\frac{5}{16.21}+...+\frac{5}{91.96}\right)\)

\(A=5\left(\frac{1}{6}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}+\frac{1}{16}-\frac{1}{21}+...+\frac{1}{91}-\frac{1}{96}\right)\)

\(A=5\left(\frac{1}{6}-\frac{1}{96}\right)\)

\(A=5.\frac{5}{32}\)

\(A=\frac{25}{32}\)

Vậy \(A=\frac{25}{32}\)

Chúc bạn học tốt ~ 

      \(\frac{3^3}{6.11}+\frac{3^3}{11.16}+.......+\frac{3^3}{91.96}\)

\(=\frac{3^3}{5}.\left(\frac{5}{6.11}+\frac{5}{11.16}+.....+\frac{5}{91.96}\right)\)

\(=\frac{27}{5}.\left(\frac{1}{6}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}+.....+\frac{1}{91}-\frac{1}{96}\right)\)

\(=\frac{27}{5}.\left(\frac{1}{6}-\frac{1}{96}\right)\)

\(=\frac{27}{5}.\frac{5}{32}\)

\(=\frac{27}{32}\)

\(\frac{3^3}{6\cdot11}+\frac{3^3}{11\cdot16}+...+\frac{3^3}{91\cdot96}\)

\(=\frac{3^3}{5}\cdot\left(\frac{1}{6}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}+...+\frac{1}{91}-\frac{1}{96}\right)\)

\(=\frac{27}{5}\cdot\left(\frac{1}{6}-\frac{1}{96}\right)\)

\(=\frac{27}{5}\cdot\frac{5}{32}=\frac{27}{32}\)

Vậy \(\frac{3^3}{6\cdot11}+\frac{3^3}{11\cdot16}+...+\frac{3^3}{91\cdot96}=\frac{27}{32}\)

\( A=\frac{3^3}{6\cdot11}+\frac{3^3}{11\cdot16}+\frac{3^3}{16\cdot21}+....+\frac{3^3}{91\cdot96}\)

\(A=\frac{3^3}{5}\cdot\left(\frac{1}{6}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}+\frac{1}{16}-\frac{1}{21}+....+\frac{1}{91}-\frac{1}{96}\right)\)

\(A=\frac{27}{5}\cdot\left(\frac{1}{6}-\frac{1}{96}\right)\)

\(A=\frac{27}{5}\cdot\frac{5}{32}=\frac{27}{32}\)

A,B,D,E :Sai

C,G,Y :Đúng

A: Đ

B: Đ 

C: S

D: Đ

E: Đ

G: S

Y: S

_HT_  NHAAAAA

Căn bậc hai số học của 25 là 

52 x 44 + 84 x 52 + 52 - 52 x 29

= 52 x ( 44 + 84 + 1 - 29 )

= 52 x        100

= 5200

nkớ k nhoa =33

    52x44+84x52+52-52x29

=52x(44+84+1+29)

=52x100

=52 000

                                                                                      HT