\(\frac{1}{4}+\frac{1}{12}+\frac{1}{24}+\frac{1}{40}+...+\frac{1}{4900}\)

\(=\frac{1}{2}.\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{2450}\right)\)

\(=\frac{1}{2}.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{49.50}\right)\)

\(=\frac{1}{2}.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{49}-\frac{1}{50}\right)\)

\(=\frac{1}{2}.\left(1-\frac{1}{50}\right)\)

\(=\frac{1}{2}.\frac{49}{50}=\frac{49}{100}\)

\(\frac{1}{4}+\frac{1}{12}+\frac{1}{24}+\frac{1}{40}+\frac{1}{60}+\frac{1}{84}\)

\(=\frac{1}{2}\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}\right)\)

\(=\frac{1}{2}\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\right)\)

\(=\frac{1}{2}\left(1-\frac{1}{7}\right)\)

\(=\frac{1}{2}.\frac{6}{7}=\frac{3}{7}\)

Đặt \(C=\frac{1}{2}+\frac{1}{12}+\frac{1}{24}+...+\frac{1}{84}\)

\(\Rightarrow\frac{C}{2}=1+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{42}\)

\(\Rightarrow C.\frac{1}{2}=1+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{6.7}\)

\(\Rightarrow C.\frac{1}{2}=1+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{6}-\frac{1}{7}\)

\(\Rightarrow C.\frac{1}{2}=1+\frac{1}{2}-\frac{1}{7}\)

\(\Rightarrow C=\left(1+\frac{1}{2}-\frac{1}{7}\right).2\)

\(\frac{3}{7}\)

Tính nhanh : 

\(A=\frac{1}{4}+\frac{1}{12}+\frac{1}{24}+\frac{1}{40}+\frac{1}{60}+\frac{1}{84}\)

\(A=2\left(\frac{1}{2\cdot4}+\frac{1}{4\cdot6}+\frac{1}{6\cdot8}+\frac{1}{8\cdot10}+\frac{1}{10\cdot12}+\frac{1}{12\cdot14}\right)\)

\(A=2\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{10}-\frac{1}{12}+\frac{1}{12}-\frac{1}{14}\right)\)

\(A=2\left(\frac{1}{2}-\frac{1}{14}\right)\)

\(A=2\cdot\frac{3}{7}\)

\(A=\frac{6}{7}\)

\(A=\frac{1}{4}+\frac{1}{12}+\frac{1}{24}+\frac{1}{40}+\frac{1}{60}+\frac{1}{84}\)

\(A=\frac{2}{8}+\frac{2}{24}+\frac{2}{48}+\frac{2}{80}+\frac{2}{120}+\frac{2}{168}\)

\(A=\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+\frac{2}{8.10}+\frac{2}{10.12}+\frac{2}{12.14}\)

\(A=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+\frac{1}{8}-\frac{1}{10}+\frac{1}{10}-\frac{1}{12}+\frac{1}{12}-\frac{1}{14}\)

\(A=\frac{1}{2}-\frac{1}{14}\)

\(A=\frac{3}{7}\)

_Chúc bạn học tốt_

A=1/4+1/12+1/24+1/40+1/60+1/84

A=2/8+2/24+2/48+2/80+2/120+2/168

A=1/2-1/4+1/4-1/6+1/6-1/8+1/8-1/10+1/10-1/12+1/12-1/14

A=1/2-1/14

A=7/14-1/14

A=6/14

A=3/7

mik ko hieu lam

\(\frac{1}{4}+\frac{1}{12}+\frac{1}{24}+\frac{1}{40}+...+\frac{1}{4900}\)

\(=\frac{1}{2}.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{49.50}\right)\)

\(=\frac{1}{2}.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{49}-\frac{1}{50}\right)\)

\(=\frac{1}{2}.\left(1-\frac{1}{50}\right)\)

\(=\frac{1}{2}.\frac{49}{50}=\frac{49}{100}\)

A = 1/4 x 8 + 1/8 x 12 + 1/12 x 16 + ... + 1/176 x 180

=> 4A = 4/4 x 8 + 4/8 x 12 + 4/12 x 16 + ... + 4/176 x 180

=> 4A = 1/4 - 1/8 + 1/8 - 1/12 + 1/12 - 1/16 + ... 1/176 - 1/180

=> 4A = 1/4 - 1/180

=> 4A = 45/180 - 1/180

=> 4A = 44/180

=> 4A = 11/45

=> A = 11/45 : 4

=> A  = 11/180

Vậy A = 11/180

A = \(\dfrac{1}{4\times8}\) + \(\dfrac{1}{8\times12}\) + \(\dfrac{1}{12\times16}\) +...+ \(\dfrac{1}{176\times180}\)

A = \(\dfrac{1}{4}\) \(\times\)( \(\dfrac{4}{4\times8}\)+ \(\dfrac{4}{12\times16}\)+...+ \(\dfrac{4}{176\times180}\))

A = \(\dfrac{1}{4}\) \(\times\)( \(\dfrac{1}{4}\) - \(\dfrac{1}{8}\) + \(\dfrac{1}{12}\) - \(\dfrac{1}{16}\) +...+ \(\dfrac{1}{176}\) - \(\dfrac{1}{180}\))

A = \(\dfrac{1}{4}\) \(\times\)(\(\dfrac{1}{4}\) - \(\dfrac{1}{180}\))

A = \(\dfrac{1}{4}\) \(\times\)\(\dfrac{11}{45}\)

A = \(\dfrac{11}{180}\)

\(\frac{1}{2x^2+10x+12}+\frac{1}{2x^2+14x+24}+\frac{1}{2x^2+18x+40}+\frac{1}{2x^2+22x+60}=\frac{1}{8}\)

<=> \(\frac{1}{2x^2+6x+4x+12}+\frac{1}{2x^2+6x+8x+24}+\frac{1}{2x^2+8x+10x+40}+\frac{1}{2x^2+12x+10x+60}=\frac{1}{8}\)

<=> \(\frac{1}{2x\left(x+3\right)+4\left(x+3\right)}+\frac{1}{2x\left(x+3\right)+8\left(x+3\right)}+\frac{1}{2x\left(x+4\right)+10\left(x+4\right)}+\frac{1}{2x\left(x+6\right)+10\left(x+6\right)}=\frac{1}{8}\)

<=> \(\frac{1}{\left(x+3\right)\left(2x+4\right)}+\frac{1}{\left(x+3\right)\left(2x+8\right)}+\frac{1}{\left(x+4\right)\left(2x+10\right)}+\frac{1}{\left(x+6\right)\left(2x+10\right)}=\frac{1}{8}\)

<=> \(\frac{1}{2\left(x+2\right)\left(x+3\right)}+\frac{1}{2\left(x+3\right)\left(x+4\right)}+\frac{1}{2\left(x+4\right)\left(x+5\right)}+\frac{1}{2\left(x+5\right)\left(x+6\right)}=\frac{1}{8}\)

<=> \(\frac{1}{2}.\left[\frac{1}{\left(x+2\right)\left(x+3\right)}+\frac{1}{\left(x+3\right)\left(x+4\right)}+\frac{1}{\left(x+4\right)\left(x+5\right)}+\frac{1}{\left(x+5\right)\left(x+6\right)}\right]=\frac{1}{8}\)

<=> \(\frac{1}{x+2}-\frac{1}{x+3}+\frac{1}{x+3}-\frac{1}{x+4}+\frac{1}{x+4}-\frac{1}{x+5}+\frac{1}{x+5}-\frac{1}{x+6}=\frac{1}{8}:\frac{1}{2}\)

<=> \(\frac{1}{x+2}-\frac{1}{x+6}=\frac{1}{4}\)

<=> \(\frac{4\left(x+6\right)-4\left(x+2\right)}{4\left(x+2\right)\left(x+6\right)}=\frac{\left(x+2\right)\left(x+6\right)}{4\left(x+2\right)\left(x+6\right)}\)

<=> \(4\left(x+6\right)-4\left(x+2\right)=\left(x+2\right)\left(x+6\right)\)

<=> \(4\left(x+6-x-2\right)=x^2+8x+12\)

<=> \(4.4=x^2+8x+12\)

<=> \(x^2+8x-4=0\)

<=> ...

Đến đây bạn tự giải tiếp. Mình bấm máy 570ES PLUS II thì ra nghiệm \(x\approx0,47\).

\(-\frac{125}{668994}\)nha

cai nay ban cu tinh nhu binh thuong thoi

sai r nha bạn