\(=\dfrac{1}{2}-\left(\dfrac{1}{3\cdot7}+\dfrac{1}{7\cdot11}+...+\dfrac{1}{23\cdot27}\right)\)

\(=\dfrac{1}{2}-\dfrac{1}{4}\left(\dfrac{4}{3\cdot7}+\dfrac{4}{7\cdot11}+...+\dfrac{4}{23\cdot27}\right)\)

\(=\dfrac{1}{2}-\dfrac{1}{4}\left(\dfrac{1}{3}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{11}+...+\dfrac{1}{23}-\dfrac{1}{27}\right)\)

\(=\dfrac{1}{2}-\dfrac{1}{4}\cdot\dfrac{9-1}{27}\)

\(=\dfrac{1}{2}-\dfrac{1}{4}\cdot\dfrac{8}{27}=\dfrac{1}{2}-\dfrac{2}{27}=\dfrac{27-4}{54}=\dfrac{23}{54}\)

2: \(=\dfrac{-2}{75}+\dfrac{5}{39}=\dfrac{33}{325}\)

3: \(=\dfrac{6}{11}\left(\dfrac{4}{9}+\dfrac{5}{9}\right)=\dfrac{6}{11}\)

4: \(=\dfrac{7}{19}\left(\dfrac{5}{13}+\dfrac{8}{13}-1\right)=-2\cdot\dfrac{7}{19}=-\dfrac{14}{19}\)

5: \(=\dfrac{2}{7}\left(\dfrac{4}{23}-\dfrac{27}{23}+1\right)=0\)

6: \(=\dfrac{3}{8}\left(\dfrac{3}{7}+\dfrac{4}{7}\right)+\dfrac{11}{8}=\dfrac{3}{8}+\dfrac{11}{8}=\dfrac{14}{8}=\dfrac{7}{4}\)

\(A=\frac{1}{1\cdot2}+\frac{2}{2\cdot4}+\frac{3}{4\cdot7}+\frac{4}{7\cdot11}+...+\frac{10}{46\cdot56}\)

\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{46}-\frac{1}{56}\)

\(A=1-\frac{1}{56}\)

\(A=\frac{55}{56}\)

\(B=\frac{4}{3\cdot7}+\frac{4}{7\cdot11}+\frac{4}{11\cdot15}+...+\frac{4}{23\cdot27}\)

\(B=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{15}+...+\frac{1}{23}-\frac{1}{27}\)

\(B=\frac{1}{3}-\frac{1}{27}\)

\(B=\frac{8}{27}\)

\(C=\frac{4}{3\cdot6}+\frac{4}{6\cdot9}+\frac{4}{9\cdot12}+...+\frac{4}{99\cdot102}\)

\(C=\frac{4}{3}\left(\frac{3}{3\cdot6}+\frac{3}{6\cdot9}+\frac{3}{9\cdot12}+...+\frac{3}{99\cdot102}\right)\)

\(C=\frac{4}{3}\left(\frac{1}{3}-\frac{1}{6}+\frac{1}{6}-\frac{1}{9}+\frac{1}{9}-\frac{1}{12}+...+\frac{1}{99}-\frac{1}{102}\right)\)

\(C=\frac{4}{3}\left(\frac{1}{3}-\frac{1}{102}\right)\)

\(C=\frac{4}{3}\cdot\frac{33}{102}\)

\(C=\frac{22}{51}\)

Các bạn giải giúp mình nha😐

Tổng các số ở nhóm chứa số 199 là :

\(\left[\left(199-1\right):2+1\right]\left(1+199\right):2=100.200:2=10000\)

Đáp số...

a)\(\frac{4}{3.7}+\frac{4}{7.11}+...+\frac{4}{23.27}=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{23}-\frac{1}{27}=\frac{1}{3}-\frac{1}{27}=\frac{8}{27}\)

b)\(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{6.7}=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{6}-\frac{1}{7}=\frac{1}{2}-\frac{1}{7}=\frac{5}{14}\)

c)\(\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{11.13}+\frac{2}{1.2}+\frac{2}{2.3}+...+\frac{2}{9.10}=\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{11}-\frac{1}{13}\right)+2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}\right)\)

\(=\frac{1}{3}-\frac{1}{13}+2\left(1-\frac{1}{10}\right)=\frac{10}{39}+\frac{9}{5}=\frac{401}{195}\)

\(1=1^3\)

\(3+5=8=2^3\)

\(7+9+11=27=3^3\)

\(13+15+17+19=64=4^3\)

\(21+23+25+27+29=125=5^3\)

Bài 1:

a; \(\dfrac{5}{18}\) + \(\dfrac{8}{19}\) - \(\dfrac{7}{21}\) + (- \(\dfrac{10}{36}\) + \(\dfrac{11}{19}\)  + \(\dfrac{1}{3}\)) - \(\dfrac{5}{8}\)

=  \(\dfrac{5}{18}\) + \(\dfrac{8}{19}\) - \(\dfrac{1}{3}\) -\(\dfrac{10}{36}\) + \(\dfrac{11}{19}\) + \(\dfrac{1}{3}\) - \(\dfrac{5}{8}\)

= (\(\dfrac{5}{18}\) - \(\dfrac{10}{36}\)) + (\(\dfrac{8}{19}\) + \(\dfrac{11}{19}\)) - (\(\dfrac{1}{3}\) - \(\dfrac{1}{3}\)) - \(\dfrac{5}{8}\)

= (\(\dfrac{5}{18}\) - \(\dfrac{5}{18}\)) + \(\dfrac{19}{19}\) - 0 - \(\dfrac{5}{8}\)

= 0 + 1 - \(\dfrac{5}{8}\)

= \(\dfrac{3}{8}\)

b; \(\dfrac{1}{13}\) + (\(\dfrac{-5}{18}\) - \(\dfrac{1}{13}\) + \(\dfrac{12}{17}\)) - (\(\dfrac{12}{17}\) - \(\dfrac{5}{18}\) + \(\dfrac{7}{5}\))

= \(\dfrac{1}{13}\) - \(\dfrac{5}{18}\) - \(\dfrac{1}{13}\) + \(\dfrac{12}{17}\) - \(\dfrac{12}{17}\) + \(\dfrac{5}{18}\) - \(\dfrac{7}{5}\)

= (\(\dfrac{1}{13}\) - \(\dfrac{1}{13}\)) + (\(\dfrac{12}{17}\) - \(\dfrac{12}{17}\)) + (-\(\dfrac{5}{18}\) + \(\dfrac{5}{18}\)) - \(\dfrac{7}{5}\)

= 0 + 0 + 0 - \(\dfrac{7}{5}\)

= - \(\dfrac{7}{5}\)

Bài 1 c;

   \(\dfrac{15}{14}\) - (\(\dfrac{17}{23}\) - \(\dfrac{80}{87}\) + \(\dfrac{5}{4}\)) + (\(\dfrac{17}{23}\) - \(\dfrac{15}{14}\) + \(\dfrac{1}{4}\))

=  \(\dfrac{15}{14}\) - \(\dfrac{17}{23}\) + \(\dfrac{80}{87}\) - \(\dfrac{5}{4}\) + \(\dfrac{17}{23}\) - \(\dfrac{15}{14}\) + \(\dfrac{1}{4}\)

= (\(\dfrac{15}{14}-\dfrac{15}{14}\)) + (\(-\dfrac{17}{23}+\dfrac{17}{23}\)) - (\(\dfrac{5}{4}\) - \(\dfrac{1}{4}\)) + \(\dfrac{80}{87}\)

= 0 + 0 - 1 + \(\dfrac{80}{87}\)

= - \(\dfrac{7}{87}\)

Bài 1 d; 

   \(\dfrac{1}{25}\) - \(\dfrac{4}{27}\) + (-\(\dfrac{23}{27}\) + \(\dfrac{-1}{25}\) - \(\dfrac{5}{43}\)) + \(\dfrac{5}{43}\) - \(\dfrac{4}{7}\)

= \(\dfrac{1}{25}\) - \(\dfrac{4}{27}\) - \(\dfrac{23}{27}\) - \(\dfrac{1}{25}\) - \(\dfrac{5}{43}\) + \(\dfrac{5}{43}\) - \(\dfrac{4}{7}\)

= (\(\dfrac{1}{25}\) - \(\dfrac{1}{25}\)) - (\(\dfrac{4}{27}\) + \(\dfrac{23}{27}\)) - (\(\dfrac{5}{43}\) - \(\dfrac{5}{43}\)) - \(\dfrac{4}{7}\)

= 0 - 1 - 0 - \(\dfrac{4}{7}\)

= \(\dfrac{-11}{7}\)

a) Ta có:

C = 5/18 + 8/19 - 7/21 + (-10/36 + 11/19 + 1/3) - 5/8

C = 5/18 + 8/19 - 1/3 - 5/18 + 11/19 + 1/3 - 5/8

C = (5/18 - 5/18) + (8/19 + 11/19) - (1/3  - 1/3) - 5/8

C = 1 - 5/8

c = 3/8

b) F = 15/14 - (17/23 - 80/87 + 5/4) + (17/23 - 15/14 + 1/4)

F = 15/14 - 17/23 + 80/87 - 5/4 + 17/23 - 15/14 + 1/4

F = (15/14 - 15/14) - (17/23 - 17/23) + 80/87 - (5/4 - 1/4)

F = 80/87 - 1

F = -7/87

c) G = 1/25 - 4/27 + (-23/27 + -1/25 - 5/43) + 5/43 - 4/7

G = 1/25 - 4/27 - 23/27 - 1/25 - 5/43 + 5/43 - 4/7

G = (1/25 - 1/25) - (4/27 + 23/27) - (5/43 - 5/43) - 4/7

G = -1 - 4/7 = -11/7