Tk mình đi mọi người mình bị âm nè!

ai tk mình mình tk lại cho!!!

trả lời câu hỏi của mình đã

bạn Hà Chí Dương phải  giúp đỡ các bạn trả lời trên Online Math đừng có dụ dỗ các bạn tk mk như vậy

a) F = \(\frac{1}{25.27}+\frac{1}{27.29}+\frac{1}{29.31}+...+\frac{1}{73.75}\)

F = \(\frac{1}{2}.\left(\frac{1}{25}-\frac{1}{27}\right)+\frac{1}{2}.\left(\frac{1}{27}-\frac{1}{29}\right)+\frac{1}{2}.\left(\frac{1}{29}-\frac{1}{31}\right)+...+\frac{1}{2}.\left(\frac{1}{73}-\frac{1}{75}\right)\)

F = \(\frac{1}{2}.\left(\frac{1}{25}-\frac{1}{27}+\frac{1}{27}-\frac{1}{29}+\frac{1}{29}-\frac{1}{31}+...+\frac{1}{73}-\frac{1}{75}\right)\)

F = \(\frac{1}{2}.\left(\frac{1}{25}-\frac{1}{75}\right)\)

F = \(\frac{1}{2}.\frac{2}{75}\)

F = \(\frac{1}{75}\)

b) G = \(\frac{15}{90.94}+\frac{15}{94.98}+\frac{15}{98.102}+...+\frac{15}{146.150}\)

G = \(\frac{15}{4}.\frac{4}{90.94}+\frac{15}{4}.\frac{4}{94.98}+\frac{15}{4}.\frac{4}{98.102}+...+\frac{15}{4}.\frac{4}{146.150}\)

G = \(\frac{15}{4}.\left(\frac{1}{90}-\frac{1}{94}\right)+\frac{15}{4}.\left(\frac{1}{94}-\frac{1}{98}\right)+\frac{15}{4}.\left(\frac{1}{98}-\frac{1}{102}\right)+...+\frac{15}{4}.\left(\frac{1}{146}-\frac{1}{150}\right)\)

G = \(\frac{15}{4}.\left(\frac{1}{90}-\frac{1}{94}+\frac{1}{94}-\frac{1}{98}+\frac{1}{98}-\frac{1}{102}+...+\frac{1}{146}-\frac{1}{150}\right)\)

G = \(\frac{15}{4}.\left(\frac{1}{90}-\frac{1}{150}\right)\)

G = \(\frac{15}{4}.\frac{1}{225}\)

G = \(\frac{1}{60}\)

<br class="Apple-interchange-newline"><div id="inner-editor"></div>12.4 +14.6 +...+198.100 

=12 (22.4 +24.6 +...+298.100 )

<br class="Apple-interchange-newline"><div id="inner-editor"></div>=12 (12 −14 +14 −16 +...+198 −1100 )

<br class="Apple-interchange-newline"><div id="inner-editor"></div>=12 (12 −14 +14 −16 +...+198 −1100 )

<br class="Apple-interchange-newline"><div id="inner-editor"></div>=12 (12 −1100 )=12 .49100 =49200 

1056 +10140 +10260 +...+101400 =53 (

c) H = \(\frac{10}{56}+\frac{10}{140}+\frac{10}{260}+...+\frac{10}{1400}\)

H = \(\frac{5}{28}+\frac{5}{70}+\frac{5}{130}+...+\frac{5}{700}\)

H = \(\frac{5}{3}.\frac{3}{28}+\frac{5}{3}.\frac{3}{70}+\frac{5}{3}.\frac{3}{130}+...+\frac{5}{3}.\frac{3}{700}\)

H = \(\frac{5}{3}.\left(\frac{3}{28}+\frac{3}{70}+\frac{3}{130}+...+\frac{3}{700}\right)\)

H = \(\frac{5}{3}.\left(\frac{3}{4.7}+\frac{3}{7.10}+\frac{3}{10.13}+...+\frac{3}{25.28}\right)\)

H = \(\frac{5}{3}.\left(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+...+\frac{1}{25}-\frac{1}{28}\right)\)

H = \(\frac{5}{3}.\left(\frac{1}{4}-\frac{1}{28}\right)\)

H = \(\frac{5}{3}.\frac{3}{14}\)

H = \(\frac{5}{14}\)

\(A=\dfrac{1}{10}+\dfrac{1}{15}+\dfrac{1}{21}+...+\dfrac{1}{120}\)

\(A=\dfrac{2}{20}+\dfrac{2}{30}+\dfrac{2}{42}+...+\dfrac{2}{240}\)

\(A=\dfrac{2}{4.5}+\dfrac{2}{5.6}+\dfrac{2}{6.7}+...+\dfrac{2}{15.16}\)

\(A=2\left(\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+...+\dfrac{1}{15.16}\right)\)

\(A=2\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+...+\dfrac{1}{15}-\dfrac{1}{16}\right)\)

\(A=2\left(\dfrac{1}{4}-\dfrac{1}{16}\right)\)

\(A=2.\dfrac{3}{16}\)

\(A=\dfrac{3}{8}\)

\(B=\dfrac{4}{3.7}+\dfrac{4}{7.11}+\dfrac{4}{11.15}+...+\dfrac{4}{107.111}\)

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

\(B=\dfrac{1}{3}-\dfrac{1}{111}\)

\(B=\dfrac{12}{37}\)

\(C=\dfrac{7}{10.11}+\dfrac{7}{11.12}+\dfrac{7}{12.13}+...+\dfrac{7}{69.70}\)

\(C=7\left(\dfrac{1}{10.11}+\dfrac{1}{11.12}+\dfrac{1}{12.13}+...+\dfrac{7}{69.70}\right)\)

\(C=7\left(\dfrac{1}{10}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{12}+\dfrac{1}{12}-\dfrac{1}{13}+...+\dfrac{1}{69}-\dfrac{1}{70}\right)\)

\(C=7\left(\dfrac{1}{10}-\dfrac{1}{70}\right)\)

\(C=7.\dfrac{3}{35}\)

\(C=\dfrac{3}{5}\)

`=>2B=(2)/(25.27)+(2)/(27.29)+(2)/(29.31)+....+(2)/(73.75)`

`=>2B=(1)/(25)-(1)/(27)+(1)/(27)-(1)/(29)+(1)/(29)-(1)/(31)+.....+(1)/(73)-(1)/(75)`

`=>2B=(1)/(25)-(1)/(75)`

`=>2B=(3)/(75)-(1)/(75)=(2)/(75)`

`=>B=(2)/(75):2`

`=>B=1/75`

\(B=\dfrac{1}{25.27}+\dfrac{1}{27.29}+\dfrac{1}{29.31}+...+\dfrac{1}{73.75}\)

\(\Rightarrow2B=\dfrac{2}{25.27}+\dfrac{2}{27.29}+...+\dfrac{2}{73.75}=\dfrac{1}{25}-\dfrac{1}{27}+\dfrac{1}{27}-\dfrac{1}{29}+...+\dfrac{1}{73}-\dfrac{1}{75}\)\(\Rightarrow2B=\dfrac{1}{25}-\dfrac{1}{75}=\dfrac{2}{75}\Rightarrow B=\dfrac{1}{75}\)

Ta có: \(B=\dfrac{1}{25\cdot27}+\dfrac{1}{27\cdot29}+\dfrac{1}{29\cdot31}+...+\dfrac{1}{73\cdot75}\)

\(=\dfrac{1}{2}\left(\dfrac{2}{25\cdot27}+\dfrac{2}{27\cdot29}+\dfrac{2}{29\cdot31}+...+\dfrac{2}{73\cdot75}\right)\)

\(=\dfrac{1}{2}\left(\dfrac{1}{25}-\dfrac{1}{27}+\dfrac{1}{27}-\dfrac{1}{29}+...+\dfrac{1}{73}-\dfrac{1}{75}\right)\)

\(=\dfrac{1}{2}\left(\dfrac{1}{25}-\dfrac{1}{75}\right)\)

\(=\dfrac{1}{2}\cdot\dfrac{2}{75}=\dfrac{1}{75}\)

\(A=7\left(\dfrac{1}{10}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{12}+...+\dfrac{1}{69}-\dfrac{1}{70}\right)\)

\(=7\left(\dfrac{1}{10}-\dfrac{1}{70}\right)=\dfrac{7.60}{700}=\dfrac{420}{700}=\dfrac{3}{5}\)

\(B=\dfrac{1}{2}\left(\dfrac{1}{25}-\dfrac{1}{27}+\dfrac{1}{27}-\dfrac{1}{29}+...+\dfrac{1}{73}-\dfrac{1}{75}\right)\)

\(=\dfrac{1}{2}\left(\dfrac{1}{25}-\dfrac{1}{75}\right)=\dfrac{1}{75}\)

\(E=\frac{1}{25\cdot27}+\frac{1}{27\cdot29}+...+\frac{1}{73\cdot75}\)

\(E=\frac{1}{2}\left(\frac{1}{25}-\frac{1}{27}+\frac{1}{27}-\frac{1}{29}+...+\frac{1}{73}-\frac{1}{75}\right)\)

\(\Rightarrow E=\frac{1}{2}\left(\frac{1}{25}-\frac{1}{75}\right)=\frac{1}{2}\cdot\frac{2}{75}=\frac{1}{75}\)

\(F=\frac{15}{90\cdot94}+\frac{15}{94\cdot98}+...+\frac{15}{146\cdot150}\)

\(F=\frac{15}{4}\cdot\left(\frac{1}{90}-\frac{1}{94}+\frac{1}{94}-\frac{1}{98}+...+\frac{1}{146}-\frac{1}{150}\right)\)

\(\Rightarrow F=\frac{15}{4}\cdot\left(\frac{1}{90}-\frac{1}{150}\right)=\frac{15}{4}\cdot\frac{1}{225}=\frac{1}{60}\)

\(G=\frac{10}{56}+\frac{10}{140}+\frac{10}{260}+...+\frac{10}{1400}\)

\(G=\frac{5}{28}+\frac{5}{70}+\frac{5}{130}+...+\frac{5}{700}\)

\(G=\frac{5}{4\cdot7}+\frac{5}{7\cdot10}+\frac{5}{10\cdot13}+...+\frac{5}{25\cdot28}\)

\(G=\frac{5}{3}\left(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+...+\frac{1}{25}-\frac{1}{28}\right)\)

\(\Rightarrow G=\frac{5}{3}\left(\frac{1}{4}-\frac{1}{28}\right)=\frac{5}{3}\cdot\frac{3}{14}=\frac{5}{14}\)

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