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

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

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

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

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

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

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

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

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

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

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

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

Ax1/2=1/2+1/6+1/12+1/20+1/30+1/42

Ax1/2=1/1.2+1/2.3+1/3.4+...+1/6.7

Ax1/2=1-1/2+1/2-1/3+1/3-1/4+....+1/6-1/7

Ax1/2=1-1/7=6/7

=>A=6/7x2=12/7

A= \(\frac{1}{4}\)+ \(\frac{1}{12}\)+ \(\frac{1}{24}\)+ \(\frac{1}{40}\)+ \(\frac{1}{60}\)+ \(\frac{1}{84}\)
2A=\(\frac{1}{2}\)+ \(\frac{1}{6}\)+ \(\frac{1}{12}\)+\(\frac{1}{20}\)+ \(\frac{1}{30}\)+ \(\frac{1}{42}\)
2A=\(\frac{1}{1.2}\)+ \(\frac{1}{2.3}\)+ \(\frac{1}{3.4}\)+ \(\frac{1}{4.5}\)+ \(\frac{1}{5.6}\)+ \(\frac{1}{6.7}\)
2A=\(1-\frac{1}{2}\)\(+\frac{1}{2}\)\(-\frac{1}{3}\)\(+...+\)\(\frac{1}{6}\)\(-\frac{1}{7}\)
2A=\(1-\frac{1}{7}\)
2A=\(\frac{6}{7}\)
A=\(\frac{6}{7}:2\)
\(\Rightarrow A=\frac{3}{7}\)
Vậy \(A=\frac{3}{7}\)
Chúc bạn học tốt nhé!

Ta có: \(B=\frac{1}{112}-\frac{1}{84}-\frac{1}{60}-\frac{1}{40}-\frac{1}{24}-\frac{1}{12}-\frac{1}{4}\)

\(\Rightarrow2B=\frac{1}{56}-\frac{1}{42}-\frac{1}{30}-\frac{1}{20}-\frac{1}{12}-\frac{1}{6}-\frac{1}{2}\)

\(\Rightarrow2B=\frac{1}{56}-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}\right)\)

\(\Rightarrow2B=\frac{1}{56}-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}\right)\)

\(\Rightarrow2B=\frac{1}{56}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\right)\)

\(\Rightarrow2B=\frac{1}{56}-\left(1-\frac{1}{7}\right)\)

\(\Rightarrow2B=\frac{1}{56}-\frac{6}{7}\)

\(\Rightarrow2B=-\frac{47}{56}\)

\(\Rightarrow B=-\frac{47}{112}\)

Hok tốt nha^^

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

CÒN CÁCH LÀM ĐANG CHƯA BIẾT

Trả lời:

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

Hmm chứ ko phải là cứ cộng hết vào là đc ạ hay phải tính nhanh?

:p

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

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

\(A=\frac{2}{2\times4}+\frac{2}{6\times8}+\frac{2}{8\times10}+\frac{2}{10\times12}+\frac{2}{12\times24}\)

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

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

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

Vậy : \(\frac{1}{4}+\frac{1}{12}+\frac{1}{24}+\frac{1}{40}+\frac{1}{60}+\frac{1}{84}=\frac{3}{7}\)\(.\)

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

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

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

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

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

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

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

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

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

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

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

<=> \(4.4=x^2+8x+12\)

<=> \(x^2+8x-4=0\)

<=> ...

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

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

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

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

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

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

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

Đặt tổng trên = A

Có : A = 1/1.2.3 + 1/2.3.4 + ...... + 1/9.10.11

2A = 2/1.2.3 + 2/2.3.4 + ...... + 2/9.10.11

     = 1/1.2 - 1/2.3 + 1/2.3 - 1/3.4 + ....... + 1/9.10 - 1/10.11

     = 1/1.2 - 1/10.11

     = 1/2 - 1/110 = 27/55

=> A = 27/55 : 2 = 27/110

Tk mk nha