\(=\frac{4+9}{4.9}-\frac{14-9}{9.14}-\frac{14+19}{14.19}+\frac{19+24}{19.24}\)

\(=\frac{4}{4.9}+\frac{9}{4.9}-\frac{14}{9.14}-\frac{9}{9.14}-\frac{14}{14.19}+\frac{19}{14.19}+\frac{19}{19.24}+\frac{24}{19.24}\)

\(\frac{1}{9}+\frac{1}{4}-\frac{1}{9}-\frac{1}{14}-\frac{1}{19}+\frac{1}{14}+\frac{1}{19}+\frac{1}{24}\)

\(=\frac{1}{4}+\frac{1}{24}=\frac{7}{24}\)

AI THẤY ĐÚNG ỦNG HỘ  NHA

Đặt \(A=\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+...+\frac{1}{21.22}\)

\(A=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+...+\frac{1}{21}-\frac{1}{22}\)

\(A=\frac{1}{5}-\frac{1}{22}\)

\(A=\frac{17}{110}\)

\(\frac{1}{5.6}\)+\(\frac{1}{6.7}\)+...+\(\frac{1}{21.22}\)

=\(\frac{1}{5}\)-\(\frac{1}{6}\)+\(\frac{1}{6}\)-\(\frac{1}{7}\)+...+\(\frac{1}{21}\)-\(\frac{1}{22}\)

=\(\frac{1}{5}\)-\(\frac{1}{22}\)

=\(\frac{17}{110}\)

\(A=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{21}-\frac{1}{22}\)

\(A=\frac{1}{5}-\frac{1}{22}\)

\(A=\frac{22}{110}-\frac{5}{110}\)

\(A=\frac{17}{110}\)

xét A và B có :

\(\frac{42}{47}\)<\(\frac{42}{45}\) (1)

theo tính chất bắc cầu ta có ;

\(\frac{37}{51}\)+\(\frac{14}{51}\)=1        ;         \(\frac{29}{37}\)+\(\frac{8}{37}\)=1  

\(\frac{31}{35}\)+\(\frac{4}{35}\)=1          ;          \(\frac{49}{63}\)+\(\frac{14}{63}\)=1

Mà \(\frac{14}{51}\)>\(\frac{14}{63}\)=> \(\frac{37}{51}\)< \(\frac{49}{63}\)(2)

ta lại có :  \(\frac{4}{35}\)=\(\frac{8}{70}\)( nhân cả tử và mẫu vs 2 )

mà \(\frac{8}{70}\)<\(\frac{8}{37}\)nên \(\frac{4}{35}\)<\(\frac{8}{37}\)=>\(\frac{29}{37}< \frac{31}{35}\)(3)

Từ (1) ; (2);(3)=>\(\frac{42}{47}+\frac{37}{51}+\frac{29}{37}< \frac{42}{45}+\frac{49}{63}+\frac{31}{35}\)

\(B=\frac{1}{90}-\frac{1}{72}-\frac{1}{56}-\frac{1}{42}-...-\frac{1}{6}-\frac{1}{2}\)

\(-B=\frac{1}{90}+\frac{1}{72}+\frac{1}{56}+...+\frac{1}{6}+\frac{1}{2}\)

\(-B=\frac{1}{10.9}+\frac{1}{9.8}+\frac{1}{8.7}+...+\frac{1}{3.2}+\frac{1}{2.1}\)

\(-B=\frac{1}{10}-\frac{1}{9}+\frac{1}{9}-\frac{1}{8}+...+\frac{1}{2}-1\)

\(-B=\frac{1}{10}-1\)

\(-B=\frac{9}{10}\)

=> \(B=\frac{-9}{10}\)

\(B=\frac{1}{90}-\frac{1}{72}-\frac{1}{56}-...-\frac{1}{6}-\frac{1}{2}\)

\(=\frac{1}{90}-\left(\frac{1}{72}+\frac{1}{56}+...+\frac{1}{6}+\frac{1}{2}\right)\)

\(=\frac{1}{90}-\left(\frac{1}{2}+\frac{1}{6}+...+\frac{1}{56}+\frac{1}{72}\right)\)

\(=\frac{1}{90}-\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{7.8}+\frac{1}{8.9}\right)\)

\(=\frac{1}{90}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}\right)\)

\(=\frac{1}{90}-\left(1-\frac{1}{9}\right)\)

\(=\frac{1}{90}-\frac{8}{9}\)

\(=-\frac{79}{90}\)

tại sao xyz ko gộp

\(=\frac{8}{9}+\frac{1}{2}-\left(\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}\right)\)

\(=\frac{8}{9}+\frac{1}{2}-\left(\frac{1}{3}-\frac{1}{9}\right)=1+\frac{1}{2}-\frac{1}{3}=1\frac{1}{6}\)

A = \(\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}\)=\(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\)= \(\frac{1}{3}-\frac{1}{10}=\frac{7}{30}\)

\(A=\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{9.10}\)

\(A=\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{9}-\frac{1}{10}\)\

\(A=\frac{1}{3}-\frac{1}{10}\)

\(A=\frac{7}{30}\)

ra la 7/30

1. Tìm x

\(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{97.100}=x\)

\(\Rightarrow\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{97}-\frac{1}{100}=x\)

\(\Rightarrow1-\frac{1}{100}=x\)

\(\Rightarrow x=\frac{99}{100}\)

\(2.Tính\)

\(\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}+\frac{1}{110}+\frac{1}{132}\)

\(=\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}+\frac{1}{10.11}+\frac{1}{11.12}\)

\(=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}+\frac{1}{11}-\frac{1}{12}\)

\(=\frac{1}{5}-\frac{1}{12}=\frac{7}{60}\)

học vui!!

Xin lỗi nha. Bài 1 mk làm sai. Lại nè:

\(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{97.100}=x\)

\(\frac{1}{3}\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{97.100}\right)=x\)

\(\frac{1}{3}.\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{100}\right)=x\)

\(\frac{1}{3}.\left(1-\frac{1}{100}\right)=x\)

\(\frac{1}{3}\cdot\frac{99}{100}=x\)

\(\frac{33}{100}=x\)

1. Tìm x

\(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+\frac{1}{10.13}+...+\frac{1}{97.100}=x\)

\(\frac{1}{3}.\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+\frac{3}{10.13}+...+\frac{3}{97.100}\right)=x\)

\(\frac{1}{3}.\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+...+\frac{1}{97}-\frac{1}{100}\right)=x\)

\(\frac{1}{3}.\left(1-\frac{1}{100}\right)=x\)

\(\frac{1}{3}.\frac{99}{100}=x\)

\(\frac{1.99}{3.100}=\frac{1.33}{1.100}=x\)

\(\frac{33}{100}=x\)

Vậy :  \(x=\frac{33}{100}\)

2 . tính

\(\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}+\frac{1}{110}+\frac{1}{132}\)

\(=\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}+\frac{1}{10.11}+\frac{1}{11.12}\)

\(=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}+\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}\)

\(=\frac{1}{5}-\frac{1}{12}=\frac{1}{5}+\frac{-1}{12}\)

\(=\frac{12}{60}+\frac{-5}{60}\)

\(=\frac{7}{60}\)

B = 1/6 + 1/12 + 1/20 + 1/30 + 1/42 + 1/56 + 1/72 + 1/90

B = \(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}\)+ \(\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}\)

B = \(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\)\(\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\)

B = \(\frac{1}{2}-\frac{1}{10}\)

B = \(\frac{2}{5}\)

B=1/6+1/12+1/20+1/30+1/42+1/56+1/72+1/90

B=1/2x3+1/3x4+1/4x5+1/5x6+1/6x7+1/7x8+1/8x9+1/9x10

B=1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6+1/6-1/7+1/7-1/8+1/8-1/9+1/9-1/10

B=1/2-1/10

B=2/5

A = 1/5.6 + 1/6.7 + 1/7.8 + 1/8.9 + 1/9.10 + 1/10.11 + 1/11.12

= 1/5 - 1/6 + 1/6 - 1/7 + 1/7 - 1/8 + ... + 1/11 - 1/12

= 1/5 - 1/12

= 12/60 - 5/60

= 7/60

Vậy A = 7/60.

Xét A , ta thấy:

\(A=\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}+\frac{1}{110}+\frac{1}{132}\)

\(A=\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}+\frac{1}{10.11}+\frac{1}{11.12}\)

Ta lại thấy: \(\frac{1}{5.6}=\frac{1}{5}-\frac{1}{6}\)

\(\frac{1}{6.7}=\frac{1}{6}-\frac{1}{7}\)

....................

\(\frac{1}{11.12}=\frac{1}{11}-\frac{1}{12}\)

\(A=\left(\frac{1}{5}-\frac{1}{6}\right)+\left(\frac{1}{6}-\frac{1}{7}\right)+....+\left(\frac{1}{11}-\frac{1}{12}\right)\)

\(A=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-....-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}\)

\(A=\frac{1}{5}+\left(-\frac{1}{6}+\frac{1}{6}\right)+\left(-\frac{1}{7}+\frac{1}{7}\right)+....+\left(-\frac{1}{11}+\frac{1}{11}\right)-\frac{1}{12}\)

\(A=\frac{1}{5}-\frac{1}{12}=\frac{7}{60}\)

\(A=\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}+\frac{1}{10.11}+\frac{1}{11.12}\)

\(A=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-...+\frac{1}{11}-\frac{1}{12}\)

\(A=\frac{1}{5}-\frac{1}{12}\)

\(A=\frac{7}{60}\)