\(A=\frac{4^2}{3.7}+\frac{4^2}{7.11}+\frac{4^2}{11.15}+...+\frac{4^2}{107.111}\)

\(A=\) \(4\left(\frac{1}{3.7}+\frac{1}{7.11}+\frac{1}{11.15}+...+\frac{1}{107.111}\right)\)

\(A=4\left(\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{15}+...+\frac{1}{107}-\frac{1}{111}\right)\)

\(A=4\left(\frac{1}{3}-\frac{1}{111}\right)\)

\(A=4.\frac{12}{37}\)

\(A=\frac{48}{37}\)

lớp 6

Ta có:

\(C=\frac{2}{3.4}+\frac{2}{4.5}+...+\frac{2}{40.41}+\frac{2}{41.42}\)

\(\Rightarrow C=2.\left(\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{40.41}+\frac{1}{41.42}\right)\)

\(\Rightarrow C=2\left(\frac{4-3}{3.4}+\frac{5-4}{4.5}+...+\frac{41-40}{40.41}+\frac{42-41}{41.42}\right)\)

\(\Rightarrow C=2.\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{40}-\frac{1}{41}+\frac{1}{41}-\frac{1}{42}\right)\)

\(\Rightarrow C=2.\left(\frac{1}{3}-\frac{1}{42}\right)=\frac{13}{21}\)

\(D=\frac{4}{3.7}+\frac{4}{7.11}+\frac{4}{11.15}+...+\frac{4}{107.111}\)

\(\Rightarrow D=\frac{7-3}{3.7}+\frac{11-7}{7.11}+\frac{15-11}{11.15}+...+\frac{111-107}{107.111}\)

\(\Rightarrow D=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{15}+...+\frac{1}{107}-\frac{1}{111}=\frac{1}{3}-\frac{1}{111}=\frac{12}{37}\)\(E=\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}+\frac{1}{110}\)

\(\Rightarrow E=\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}+\frac{1}{10.11}\)

\(\Rightarrow E=\frac{5-4}{4.5}+\frac{6-5}{5.6}+\frac{7-6}{6.7}+\frac{8-7}{7.8}+\frac{9-8}{8.9}+\frac{10-9}{9.10}+\frac{11-10}{10.11}\)

\(\Rightarrow E=\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}+\frac{1}{10}-\frac{1}{11}=\frac{1}{4}-\frac{1}{11}=\frac{7}{44}\)

\(\Leftrightarrow\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{15}+...+\frac{1}{107}-\frac{1}{111}\)

\(\Rightarrow=\frac{1}{3}-\frac{1}{111}\)

\(=\frac{12}{37}\)

k nha

\(=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{15}+...+\frac{1}{107}-\frac{1}{111}\)

\(=\frac{1}{3}-\frac{1}{111}\)

\(=\frac{108}{333}=\frac{12}{37}\)

ten gia mao nobita 

nobi k the gioi nhu the 

do gia :v

Ta có A = \(\frac{4}{3.7}+\frac{4}{7.11}+..............+\frac{4}{107.111}\)

=> A = \(\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+.............+\frac{1}{107}-\frac{1}{111}\)

A = \(\frac{1}{3}-\frac{1}{111}=\frac{12}{37}\)

k nha bạn

4x(\(\frac{1}{3.7}+...+\frac{1}{107.111}\) )

4(\(\frac{1}{3}-\frac{1}{7}+...+\frac{1}{107}-\frac{1}{111}\))

4(\(\frac{1}{3}-\frac{1}{111}\))

4.\(\frac{12}{37}\)

48/37

1) =3/2 . 4/3 . 5/4 ...... 100/99 = 100/2 = 50

2) = -1/2 . -2/3 .-3/4 ..... . -98/99 = 1/99 (Tích này có 98 thừa số âm, 98 là số chẵn nên tích mang dấu dương)

3) =(1.3)/(2.2) . (2.4)/(3.3) . (3.5)/(4.4) ......... (29.31)/(30.30)

    = (1.2.3.....31)/(2.3.4......30) . (3.4.5......29)/(2.3.4....30)

    = 31 . 1/60

    = 31/60

a) 4/ 3x7 + 4/7x11+ 4/11x15+...+ 4/107x111

=1/3-1/7+ 1/7-1/11+ 1/11- 1/15+...+1/107 - 1/111

= 1/3-1/111

=12/37

\(b,\frac{3^2}{8\cdot11}+\frac{3^2}{11\cdot14}+\frac{3^2}{14\cdot17}+...+\frac{3^2}{197\cdot200}\)

\(=3\left(\frac{3}{8\cdot11}+\frac{3}{11\cdot14}+\frac{3}{14\cdot17}+...+\frac{3}{197\cdot200}\right)\)

\(=3\left(\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{17}+...+\frac{1}{197}-\frac{1}{200}\right)\)

\(=3\left(\frac{1}{8}-\frac{1}{200}\right)\)

\(=3\cdot\frac{3}{25}=\frac{9}{25}\)

2+12345678-5=

\(\frac{4}{3.7}+\frac{4}{7.11}+\frac{4}{11.15}+...+\frac{4}{3\left(x-1\right)\left(3x+3\right)}=\frac{3}{10}\)

\(\Rightarrow\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{4}{\left(3x-1\right)\left(3x+3\right)}=\frac{3}{10}\)

\(\Rightarrow\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{3x-1}-\frac{1}{3x+3}=\frac{3}{10}\)(Vì 3x + 3 lớn hơn 3x - 1 là 4 đơn vị)

\(\Rightarrow\frac{1}{3}-\frac{1}{3x+3}=\frac{3}{10}\)

\(\Rightarrow\frac{x+1-1}{3x+3}=\frac{3}{10}\)

\(\Rightarrow\frac{x}{3x+3}=\frac{3}{10}\)

\(\Rightarrow10x=3.\left(3x+3\right)\)

\(\Rightarrow10x=9x+9\)

\(\Rightarrow x=9\)

Vậy...

thanks