\(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{15.16}\)

\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{15}-\frac{1}{16}\)

\(=1-\frac{1}{16}=\frac{15}{16}\)

\(\frac{1}{1x2}+\frac{1}{2x3}+...+\frac{1}{15x16}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{15}-\frac{1}{16}\)

\(=1-\frac{1}{16}\)

\(=\frac{15}{16}\)

Cảm ơn nhé

\(1+\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2017.2018}\)

\(=1+\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2017}-\frac{1}{2018}\right)\)

\(=1+\left(1-\frac{1}{2018}\right)\)

\(=1+\left(\frac{2018}{2018}-\frac{1}{2018}\right)\)

\(=1+\left(\frac{2017}{2018}\right)\)

\(=\frac{2018}{2018}+\frac{2017}{2018}=\frac{4035}{2018}\)

\(1+\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}...+\frac{1}{2017\cdot2018}\)

\(=1+\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}...+\frac{1}{2017}-\frac{1}{2018}\right)\)

\(=1+\left(1-\frac{1}{2018}\right)\)

\(=1+\frac{2017}{2018}\)

\(=1+\frac{2017}{2018}\)

\(=\frac{4035}{2018}\)

\(1+\frac{1}{1x2}+\frac{1}{2x3}+\frac{1}{3x4}+...+\frac{1}{2017x2018}\)

\(=1+\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}+...+\frac{1}{2017}-\frac{1}{2018}\right)\)

\(=1+\left(1-\frac{1}{2018}\right)\)

\(=1+\frac{2017}{2018}\)

\(=\frac{4035}{2018}\)

Mình không thể giải thích được nhưng kết quả chắc chắn là : \(\frac{8}{9}\)

đặt A=1/1×2+1/2×3+1/3×4+

bn đặt ra đi sau đó rồi tính vd 1/1x2=1/1-1/2

\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}=1-\frac{1}{100}=\frac{99}{100}\)

Đặt \(M=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\)

\(M=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\)

\(M=1-\frac{1}{100}\)

\(M=\frac{99}{100}\)

\(\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+\frac{1}{4\times5}+....+\frac{1}{99\times100}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)

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

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

1/1×2 + 1/2×3 + 1/3×4 + 1/4×5 + ... + 1/99×100

= 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 + ... + 1/99 - 1/100

= 1 - 1/100

= 99/100

Ta có : 

\(C=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{13.14}\)

\(C=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{13}-\frac{1}{14}\)

\(C=\left(\frac{1}{2}-\frac{1}{2}\right)+\left(\frac{1}{3}-\frac{1}{3}\right)+...+\left(1-\frac{1}{14}\right)\)

\(C=1-\frac{1}{14}\)

\(C=\frac{14}{14}-\frac{1}{14}\)

\(C=\frac{14-1}{14}\)

\(C=\frac{13}{14}\)

Vậy \(C=\frac{13}{14}\)

Chúc bạn học tốt ~

\(C=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+....+\frac{1}{13\cdot14}\)

\(C=\frac{2-1}{1\cdot2}+\frac{3-2}{2\cdot3}+\frac{4-3}{3\cdot4}+....+\frac{14-13}{13\cdot14}\)

\(C=\frac{2}{1\cdot2}-\frac{1}{1\cdot2}+\frac{3}{2\cdot3}-\frac{2}{2\cdot3}+\frac{4}{3\cdot4}-\frac{3}{3\cdot4}+....+\frac{14}{13\cdot14}-\frac{13}{13\cdot14}\)

\(C=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{13}-\frac{1}{14}\)

\(C=1-\frac{1}{14}\)

\(C=\frac{13}{14}\)

dấu "." là dấu nhân nhs

Ta có : \(C=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{13.14}\)

\(\Leftrightarrow C=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{13}-\frac{1}{14}\)

\(\Leftrightarrow C=1-\frac{1}{14}=\frac{13}{14}.\)Vậy \(C=\frac{13}{14}.\)

1/1x2+1/2x3+...+1/49x50

=1-1/2+1/2-1/3+.....+1/49-1/50

=1-1/50(1)

Ta co   1(2)

So sanh (1) voi (2) ta thay 1-1/50<1

=>1/1x2+...+1/49x50<1

(Phuong phap khu)

\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{49.50}\)

=\(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{49}-\frac{1}{50}\)

=\(\frac{1}{1}-\frac{1}{50}=\frac{50}{50}-\frac{1}{50}=\frac{49}{50}<1\)

Vậy \(\frac{49}{50}<1\)

1/1x2 + 1/2x3 + 1/3x4 + 1/4x5 + 1/5x6 + 1/6x7

=1/1-1/2+1/2-1/3+...-1/7

=1+(1/2-1/2+1/3-1/3+...+1/6-1/6)-1/7

=1 +0+0+...-1/7

=1-1/7

=6/7

hình như là 8/7
 

Kếết quả là 116/24

Rút gọn ra phân số tối giản là 29/6

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2010}-\frac{1}{2011}\)

\(=1-\frac{1}{2011}\)

\(=\frac{2010}{2011}\)

=1/1-1/2+1/2-1/3+1/3-1/4+...+1/2010-1/2011

= 1 - 1/2011

= 2010/ 2011

Đáp số: 2010/2011

Chúy ý công thức: \(\frac{1}{n\left(n+1\right)}=\frac{1}{n}-\frac{1}{n+1}\)

= 1/1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/2010 - 1/2011

= 1 - 1/2011

= 2010/2011

Đáp sô: 2010/2011

Chú ý công thưc:              \(\frac{1}{n\left(n+1\right)}=\frac{1}{n}-\frac{1}{n+1}\)