\(\dfrac{1}{3}+\dfrac{3}{7}+\dfrac{1}{2\cdot7}+\dfrac{5}{2\cdot13}+\dfrac{3}{13\cdot4}+\dfrac{5}{4\cdot21}\)

\(=\dfrac{1}{3}+\dfrac{3}{7}+\dfrac{1}{14}+\dfrac{5}{84}+\dfrac{5}{26}+\dfrac{3}{42}\)

\(=\dfrac{1}{3}+\dfrac{3}{7}+\dfrac{1}{14}+\dfrac{5}{84}+\dfrac{3}{42}+\dfrac{5}{26}\)

\(=\dfrac{28+36+6+5+6}{84}+\dfrac{5}{26}\)

\(=\dfrac{5}{26}+\dfrac{81}{84}\)

\(=\dfrac{5}{26}+\dfrac{27}{28}=\dfrac{421}{364}\)

1.    \(\frac{9}{13}-\frac{11}{26}+\frac{1}{2}=\frac{10}{13}\)

2.     \(\frac{1}{5}+\frac{2}{7}+\frac{7}{35}=\frac{24}{35}\)

3.      \(\frac{4}{9}+\frac{1}{3}:\frac{3}{8}=\frac{4}{3}\)

4.     \(\frac{21}{25}-\frac{1}{5}.\frac{2}{3}=\frac{53}{75}\)

\(x-\frac{3}{4}=6\cdot\frac{3}{8}\)

\(x-\frac{3}{4}=\frac{9}{4}\)

          \(x=\frac{9}{4}+\frac{3}{4}\)

          \(x=\frac{12}{4}=3\)

\(\frac{7}{8}\cdot x=3-\frac{1}{2}\)

\(\frac{7}{8}\cdot x=\frac{5}{2}\)

         \(x=\frac{7}{8}:\frac{5}{2}\)

          \(x=\frac{7}{20}\)

\(x+\frac{1}{2}\cdot\frac{1}{3}=\frac{3}{4}\)

\(x+\frac{1}{6}=\frac{3}{4}\)

          \(x=\frac{3}{4}-\frac{1}{6}\)

          \(x=\frac{7}{12}\)

a: ta có: \(5\cdot8^{12}\cdot9^5\cdot\left(-3\right)^2\cdot\left(-16\right)^5\)

\(=5\cdot\left(2^3\right)^{12}\cdot\left(3^2\right)^5\cdot3^2\cdot\left(-1\right)\cdot2^{20}\)

\(=-5\cdot3^{12}\cdot2^{56}\)

Ta có: \(7\cdot49^3\cdot81^5\cdot\left(-7\right)^{10}\cdot\left(-8\right)^5\)

\(=-7\cdot\left(7^2\right)^3\cdot\left(3^4\right)^5\cdot7^{10}\cdot\left(2^3\right)^5\)

\(=-7\cdot7^6\cdot3^{20}\cdot7^{10}\cdot2^{15}=-7^{17}\cdot2^{15}\cdot3^{20}\)

Ta có: \(B=\frac{5\cdot8^{12}\cdot9^5\cdot\left(-3\right)^2\cdot\left(-16\right)^5}{7\cdot49^3\cdot81^5\cdot\left(-7\right)^{10}\cdot\left(-8\right)^5}\)

\(=\frac{-5\cdot3^{12}\cdot2^{56}}{-7^{17}\cdot2^{15}\cdot3^{20}}=\frac{5\cdot2^{41}}{7^{17}\cdot3^8}\)

b: \(13\cdot4^7\cdot9^{15}-9^7\cdot\left(-16\right)^4\)

\(=13\cdot2^{14}\cdot3^{20}-3^{14}\cdot2^{16}\)

\(=3^{14}\cdot2^{14}\left(13\cdot3^6-2^2\right)\)

\(15\cdot3^{12}\cdot2^8+3\cdot27^5\cdot\left(-8\right)^{10}\)

\(=3\cdot5\cdot3^{12}\cdot2^8+3\cdot\left(3^3\right)^5\cdot\left(2^3\right)^{10}=3^{13}\cdot5\cdot2^8+3^{16}\cdot2^{30}\)

\(=3^{13}\cdot2^8\left(5+3^3\cdot2^{22}\right)\)
Ta có: \(C=\frac{13\cdot4^7\cdot9^{15}-9^7\cdot\left(-16\right)^4}{15\cdot3^{12}\cdot2^8+3\cdot27^5\cdot\left(-8\right)^{10}}\)

\(=\frac{3^{14}\cdot2^{14}\left(13\cdot3^6-2^2\right)}{3^{13}\cdot2^8\left(5+3^3\cdot2^{22}\right)}=\frac{3\cdot2^6\cdot\left(13\cdot3^6-2^2\right)}{5+3^3\cdot2^{22}}\)

\(A=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-...-\frac{1}{100}=1-\frac{1}{100}=\frac{99}{100}\)

Phạm Trần Khánh An : l.i.k.e tiếp cái con khỉ

\(A=\frac{1}{2}.\frac{1}{3}+\frac{1}{3}.\frac{1}{4}+\frac{1}{4}.\frac{1}{5}+..........+\frac{1}{8}.\frac{1}{9}=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+......+\frac{1}{8.9}=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-.......+\frac{1}{8}-\frac{1}{9}=\frac{1}{2}-\frac{1}{9}=\frac{7}{18}\)

\(B=\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+....+\frac{1}{110}=\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+.....+\frac{1}{10.11}=\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-.....+\frac{1}{10}-\frac{1}{11}=\frac{1}{4}-\frac{1}{11}=\frac{7}{44}\)

\(\text{c,d cơ bản tự làm nha }\)

A=>1.1/2.3+1.1/3.4+1.1/4.5+1.1/5.6+1.11/6.7+.1/7.8+1.1/8.9

=>1/2.3+1/3.4+1/4.5+1/6.7+1/7.8+1/8.9

=>1/2-1/3-1/4-1/5-1/6-1/7-1/8-1/9

=>1/2-1/9=>9/18-2/18=>7/18

Vậy A= 7/18

a: =3/8-1/4

=3/8-2/8

=1/8

b: =-5/9+3/5-1/9+2/5

=-2/3+1

=1/3

c: =21/7*5/25=3/5

d: =3/4+11/10:(2/5-3/2)-1/9

=-13/36

1 . Cách tính số hạng 

( số cuối - số đầu ) : khoảng cách + 1

Cách tính tổng của 1 dãy :

( số đầu + số cuối ) . khoảng cách : 2

b . Dãy số đó có số số hạng là : ( 2005 - 1 ) : 2 + 1 = 1003 ( số )

Tổng của dãy số đó là : ( 2005 + 1 ) x 1003 : 2 = 1006009

3.TỔng là : ( 1024 + 2 ) x 10 : 2 = 5130

cậu diễn giải đầy đủ phần c vàb