\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}\)

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

                                                                     \(=\frac{5}{6}\)

1/1.2+1/2.3+1/3.4+1/4.5+1/5.6

=1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6

=1-1/6

=5/6

1/1 - 1/2+1/2 -1/3+1/3 - 1/4+1/4 - 1/5+1/5 - 1/6

=1-( -1/2+1/2+-1/3+1/3+-1/4+1/4+-1/5+1/5)-1/6

=1 - 1/6

=6/6 - 1/6

=5/6

\(M=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}\)

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

\(M=1-\frac{1}{7}\)

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

6/7

k cho mình nha

Kết quả:\(\frac{6}{7}\)

Đúng 100% nhé!!

~Shizadon~

mk bít lm cách lớp 5, vừa học

Cần ko bn

\(\frac{1}{9.10}-\frac{1}{8.9}-\frac{1}{7.8}-\frac{1}{6.7}-\frac{1}{5.6}-\frac{1}{4.5}-\frac{1}{3.4}-\frac{1}{2.3}-\frac{1}{1.2}\)

\(=-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}\right)+\frac{1}{9.10}\)

\(=-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{8}-\frac{1}{9}\right)+\frac{1}{90}\)

\(=-\left(1-\frac{1}{10}\right)+\frac{1}{90}\)

\(=-\frac{9}{10}+\frac{1}{90}\)

= ...

bn tự tính nha!
 

\(A=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{5\cdot6}\)

\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{5}-\frac{1}{6}\)

\(A=1-\frac{1}{6}\)

\(A=\frac{5}{6}\)

\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{5}-\frac{1}{6}\)

\(A=1-\frac{1}{6}\)

\(A=\frac{5}{6}\)

\(B=\frac{3}{2}.\frac{4}{3}.\frac{5}{4}.....\frac{100}{99}\)

\(B=\frac{100}{2}\)

\(B=\left(1+\frac{1}{2}\right)\cdot\left(1+\frac{1}{3}\right)\cdot...\cdot\left(1+\frac{1}{99}\right)\)

\(B=\frac{3}{2}\cdot\frac{4}{3}\cdot...\cdot\frac{100}{99}\)

\(B=\frac{3\cdot4\cdot...\cdot100}{2\cdot3\cdot...\cdot99}\)

\(B=\frac{100}{2}\)

\(B=50\)

\(\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\right)x<\frac{13}{7}\)

\(\left(1-\frac{1}{7}\right).x<\frac{13}{7}\)

\(\frac{6}{7}.x<\frac{13}{7}\Leftrightarrow6x<13\Leftrightarrow x<2,1\left(6\right)\)

x nguyên dương => x thuộc {1;2}

Vậy tập hợp có 2 phần tử

vay tap hop co 2 phan tu

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

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

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

Đây là tính chứ chứng minh cái gì ? 

\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{9}-\frac{1}{10}\)

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

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

Lập luận: 1/1.2 = 1/1 - 1/2 ; 1/2.3 = 1/2 - 1/3 ; 1/3.4 = 1/3 - 1/4 ; làm tương tự với các số kia.

Ta có: 1 - 1/2 + 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

= 1 - 1/10

= 9/10

\(M=\frac{1}{9.10}-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{8.9}\right)=\frac{1}{90}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{8}-\frac{1}{9}\right)=\frac{1}{90}-\left(1-\frac{1}{9}\right)=\frac{1}{90}-\frac{8}{9}=-\frac{79}{90}\)

yyyyyyyyyyyyyyyyyyyyyyyyyyyy

000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000

= \(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{21111111}-\frac{2}{21111112}\)

= \(\frac{1}{1}-\frac{1}{21111112}\)(\(-\frac{1}{2}\)rút gọn cho \(+\frac{1}{2}\)và cứ như vậy đến khi chỉ còn 2 phân số \(\frac{1}{1}\)và \(\frac{1}{21111112}\))

= \(\frac{21111111}{21111112}\)

100% đúng nha bạn

\(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{201111111}-\frac{1}{2011111112}\)

\(\frac{1}{1}-\frac{1}{201111112}\)

\(\frac{201111111}{201111112}\)

=  1 1 − 2 1 + 2 1 − 3 1 + ... + 21111111 1 − 21111112 2

=  1 1 − 21111112 1

(− 2 1 rút gọn cho + 2 1 và cứ như vậy đến khi chỉ còn 2 phân số  1 /1 và  21111112