Đặt \(K=\frac{1}{3}+\frac{1}{3^2}+.....+\frac{1}{3^{99}}\)

\(3K=1+\frac{1}{3}+\frac{1}{3^2}+.....+\frac{1}{3^{98}}\)

\(3K-K=1+\frac{1}{3}+\frac{1}{3^2}+.....+\frac{1}{3^{98}}-\left(\frac{1}{3}+\frac{1}{3^2}+.....+\frac{1}{3^{99}}\right)\)

\(2K=\)\(1-\frac{1}{3^{99}}\)

\(K=\frac{1-\frac{1}{3^{99}}}{2}\)

Có \(1-\frac{1}{3^{99}}\) < \(\frac{1}{2}\)

\(\Rightarrow K\) < \(\frac{1}{2}\)

Vậy \(\left(\frac{1}{3}+\frac{1}{3^2}+.....+\frac{1}{3^{99}}\right)\) < \(\frac{1}{2}\)

a) \(\frac{3}{4} + \left( {\frac{1}{2} - \frac{1}{3}} \right) = \frac{9}{{12}} + \left( {\frac{6}{{12}} - \frac{4}{{12}}} \right) = \frac{9}{{12}} + \frac{2}{{12}} = \frac{{11}}{{12}}\)

\(\frac{3}{4} + \frac{1}{2} - \frac{1}{3} = \frac{9}{{12}} + \frac{6}{{12}} - \frac{4}{{12}} = \frac{{15}}{{12}} - \frac{4}{{12}} = \frac{{11}}{{12}}\)

Vậy \(\frac{3}{4} + \left( {\frac{1}{2} - \frac{1}{3}} \right)\) = \(\frac{3}{4} + \frac{1}{2} - \frac{1}{3}\)    

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

 \(\frac{2}{3} - \frac{1}{2} - \frac{1}{3} = \frac{4}{6} - \frac{3}{6} - \frac{2}{6} = \frac{1}{6} - \frac{2}{6} = \frac{{ - 1}}{6}\)

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

`#3107`

`a)`

`3/4 + (1/2 - 1/3)`

`= 3/4 + (3/6 - 2/6)`

`= 3/4 + 1/6`

`= 11/12`

`3/4 + 1/2 - 1/3`

`= 9/12 + 6/12 - 4/12`

`= (9 + 6 - 4)/12`

`= 11/12`

Vì `11/12 = 11/12`

`=> 3/4 + (1/2 - 1/3) = 3/4 + 1/2 - 1/3`

`b)`

`2/3 - (1/2 + 1/3)`

`= 2/3 - (3/6 + 2/6)`

`= 2/3 - 5/6`

`= -1/6`

`2/3 - 1/2 - 1/3`

`= 4/6 - 3/6 - 2/6`

`= (4 - 3 - 2)/6`

`= -1/6`

Vì `-1/6 = -1/6`

`=> 2/3 - (1/2 + 1/3) = 2/3 - 1/2 - 1/3`

2 vế bằng nhau

100-(1+1/2+1/3+...+1/100) = 1/2+2/3+3/4+...+99/100

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

100 = 1 + 1/2 + 1/2 + 1/3 + 2/3 + ... + 1/100 + 99/100 (cùng cộng 2 vế với (- 1-1/2-1/3-...-1/100)

100 = 1 + 1 + 1 + ... + 1 (100 số hạng)

100 = 100

Vậy   100-(1+1/2+1/3+...+1/100) = 1/2+2/3+3/4+...+99/100

Cảm ơn bạn!

Cho mk sửa xíu : ( 1/2)^-1 ; (1/2)^-2

(1/2)^-1=2

(1/2)^-2=4

có 2<4

=>(1/2)^-1<(1/2)^-2

Ta có :

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

\(\left(\frac{1}{3}\right)^{-1}=3\)

\(\left(\frac{1}{2}\right)^{-2}=\left(2^{-1}\right)^{-2}=2^2=4\)

\(\left(\frac{1}{4}\right)^{-1}=\left(4^{-1}\right)^{-1}=4\)

\(\left(\frac{1}{3}\right)^{-2}=\left(3^{-1}\right)^{-2}=3^2=9\)

Do đó ta có :

\(\left(\frac{1}{2}\right)^{-1}< \left(\frac{1}{3}\right)^{-1}< \left(\frac{1}{2}\right)^{-2}=\left(\frac{1}{4}\right)^{-1}< \left(\frac{1}{3}\right)^{-2}\)

\(\frac{3}{2^2}.\frac{8}{3^2}.\frac{15}{4^2}.....\frac{899}{30^2}\)

\(=\frac{1.3}{2.2}.\frac{2.4}{3.3}.\frac{3.5}{4.4}.....\frac{29.31}{30.30}=\frac{1.2.3.....29}{2.3.4.....30}.\frac{3.4.5.....31}{2.3.4.....30}\)

\(=\frac{1}{2}.\frac{31}{30}=\frac{31}{60}\)

A=(2/3+3/4+...+99/100)x(1/2+2/3+3/4+...+98/99)-(1/2+2/3+...+99/100)x(2/3+3/4+4/5+...98/99)

ta cho nó dài hơn như sau

A=(2/3+3/4+4/5+5/6+....+98/99+99/100)

ta thấy các mẫu số và tử số giống nhau nên chệt tiêu các số

2:3:4:5...99 vậy ta còn các số 2/100

ta làm vậy với(1/2+2/3+3/4+.....+98/99) thi con 1/99

làm vậy với câu (1/2+2/3+...+99/100) thì ra la 1/100

vậy với (2/3+3/4+...+98/99) ra 2/99

xùy ra ta có 2/100.1/99-1/100.2/99=1/50x1/99-1/100x2/99=tự tinh nhe mình ngủ đây