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11 tháng 7 2016

Đặt \(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{50}}\)

\(2A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{49}}\)

=> \(A=2A-A=1-\frac{1}{2^{50}}< 1\)

=> \(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{50}}< 1\)

11 tháng 7 2016

\(\text{Đ}\text{ặt}\) \(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+.....+\frac{1}{2^{50}}\)

\(\Rightarrow2A=1+\frac{1}{2}+\frac{1}{2^2}+.....+\frac{1}{2^{49}}\)

\(\Rightarrow2A-A=A=1-\frac{1}{2^{50}}< 1\)

14 tháng 5 2018

Đặt \(A=\frac{1}{2}+\frac{1}{2^2}+......+\frac{1}{2^{50}}\)

      \(2A=1+\frac{1}{2}+...+\frac{1}{2^{49}}\)

       \(2A-A=1-\frac{1}{2^{50}}\)

     \(A=1-\frac{1}{2^{50}}< 1\)

       \(\Rightarrow A< 1\)

14 tháng 8 2016

\(S=\frac{3}{1^2\cdot2^2}+\frac{5}{2^2\cdot3^2}+\frac{7}{3^2\cdot4^2}+...+\frac{99}{49^2\cdot50^2}\)

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

\(=1-\frac{1}{50^2}=\frac{2499}{2500}\)

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

\(=\frac{1}{1\cdot3}+\frac{1}{2\cdot4}+\frac{1}{3\cdot5}+...+\frac{1}{49\cdot51}\)

\(=\frac{1}{2}\cdot\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{2}-\frac{1}{4}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{49}-\frac{1}{51}\right)\)

\(=\frac{1}{2}\cdot\left(1+\frac{1}{2}-\frac{1}{51}\right)=\frac{151}{204}\)

Vì \(\frac{2499}{2500}>\frac{151}{204}\)nên S>T

14 tháng 8 2016

JOKER_Võ Văn Quốc, T = \(\frac{1}{2}.\left(1-\frac{1}{51}+\frac{1}{2}-\frac{1}{50}\right)\)mới đúng
Sẽ dễ hơn nếu bạn chia ra 2 vế \(\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{49.51}\)và \(\frac{1}{2.4}+\frac{1}{4.6}+...+\frac{1}{48+50}\)

4 tháng 8 2018

ta có 1/2^2<1/2

        1/2^3<1/2

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

      1/2^50<1/2

\(\Rightarrow\)1/2*50>1/2^1+1/2^2+1/2^3+...........+1/2^50

\(\Rightarrow\)

4 tháng 8 2018

Tìm 2A 

Rồi lấy 2A - A là ra

Ok

\(A=\frac{1}{2}+\frac{1}{2^2}+.....+\frac{1}{2^{49}}+\frac{1}{2^{50}}\)

\(2A=2.\left(\frac{1}{2}+\frac{1}{2^2}+....+\frac{1}{2^{49}}+\frac{1}{2^{50}}\right)\)

\(2A=1+\frac{1}{2}+\frac{1}{2^2}+.....+\frac{1}{2^{49}}\)

\(2A-A=\left(1+\frac{1}{2}+\frac{1}{2^2}+....+\frac{1}{2^{48}}+\frac{1}{2^{49}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+.....+\frac{1}{2^{50}}\right)\)

\(A=1-\frac{1}{2^{50}}\)

\(\Rightarrow A< 1\)

HQ
Hà Quang Minh
Giáo viên
19 tháng 9 2023

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`

11 tháng 3 2020

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31 tháng 5 2020

? crisdevergamemer

22 tháng 9 2016

Ta có: \(\frac{1}{2^2}< \frac{1}{1.2}\)

            \(\frac{1}{3^2}< \frac{1}{2.3}\)

            ....................

             .....................

             \(\frac{1}{100^2}< \frac{1}{99.100}\)

Nên \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+......+\frac{1}{100^2}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.....+\frac{1}{99.100}\)

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

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

22 tháng 9 2016

\(\frac{99}{100}\)\(\frac{3}{4}\)thì sao mà so sánh được