@Miyuki Misaki, @Nguyễn Trúc Giang, @Nguyễn Lê Phước Thịnh, @White Hold

a, Ta có : S = \(\frac{1}{50}+\frac{1}{51}+\frac{1}{52}+...+\frac{1}{99}\)

⇔ S = \(\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{99}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{49}\right)\)

⇔ \(S=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{99}\right)-2.\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{98}\right)\)

⇔\(S=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{98}-\frac{1}{99}\) ( 99 số hạng)

⇔ S = \(\left(1-\frac{1}{2}+\frac{1}{3}\right)-\left(\frac{1}{4}-\frac{1}{5}\right)-\left(\frac{1}{6}-\frac{1}{7}\right)-...-\left(\frac{1}{98}-\frac{1}{99}\right)\)

⇔ S = \(\frac{5}{6}-\left(\frac{1}{4}-\frac{1}{5}\right)-\left(\frac{1}{6}-\frac{1}{7}\right)-...-\left(\frac{1}{98}-\frac{1}{99}\right)\)

Mà ta có \(\left(\frac{1}{4}-\frac{1}{5}\right)-\left(\frac{1}{6}-\frac{1}{7}\right)-...-\left(\frac{1}{98}-\frac{1}{99}\right)\) < 0

⇔ \(-\)\(\left(\frac{1}{4}-\frac{1}{5}\right)-\left(\frac{1}{6}-\frac{1}{7}\right)-...-\left(\frac{1}{98}-\frac{1}{99}\right)\) > 0

Như vậy ta được S > \(\frac{5}{6}\) đpcm

b, \(A=\frac{1}{10}+\frac{1}{11}+\frac{1}{12}+..+\frac{1}{99}+\frac{1}{100}\) ( 91 số hạng)

Ta có \(\frac{1}{11}>\frac{1}{100};\frac{1}{12}>\frac{1}{100};..;\frac{1}{99}>\frac{1}{100}\)

⇒ \(A>\frac{1}{10}+\frac{1}{100}+\frac{1}{100}+\frac{1}{100}+...+\frac{1}{100}+\frac{1}{100}\) (90 số hạng 100)

⇒ A \(>\frac{10}{100}+90.\frac{1}{100}\)

⇒ A > \(\frac{10}{100}+\frac{90}{100}\)

⇒ A > \(\frac{100}{100}=1\)

Vậy ...

​Ta có: 151+152+...+175>175+175+...+175=2575=13​

176+177+...+1100>1100+1100+...+1100=25100=14

​=> S>13+14=712​ (1)

Ta có: 151+152+...+175<150+150+...+150=2550=12​

176+177+...+1100<175+175+...+175=2575=13​

=> S<12+13=56 ​(2)

Từ (1) và (2) => 712​ < S<56 ​( đpcm )

Ta có:

- 1/51 > 1/75, 1/52 > 1/75 ...

=> 1/51 + 1/52 + ... + 1/75 > 1/75 + ... 1/75 = 25/75 = 1/3

- 1/76 > 1/100, 1/77 > 1/100 ...

=> 1/76 + 1/77 + ... + 1/100 > 1/100 + ... + 1/100 = 25/100 = 1/4

Từ đó : S = ( 1/51 + ... + 1/75 ) + ( 1/76 + ... + 1/100 ) > 1/3 + 1/3 = 7/12 (1)

- 1/51 < 1/50, 1/52 < 1/50 ... 

=> 1/51 + 1/52 + ... + 1/75 < 1/50 + ... 1/50 = 25/50 = 1/2

- 1/76 < 1/75, 1/77 < 1/75...

=> 1/76 + 1/77 + ... + 1/100 < 1/75 + ... + 1/75 = 25/75 = 1/3

Từ đó : S = ( 1/51 + ... + 1/75 ) + ( 1/76 + ... + 1/100 ) < 1/2 + 1/3 = 5/6 (2)

từ (1) và (2) => 5/6 > S > 7/12

* Chúc bn học tốt !!!

C>1   vì c>1

a, Ta có: \(A=\frac{1}{11}+\frac{1}{12}+...+\frac{1}{50}=\left(\frac{1}{11}+\frac{1}{12}+...+\frac{1}{30}\right)+\left(\frac{1}{31}+\frac{1}{32}+...+\frac{1}{60}\right)\)

Nhận xét: \(\frac{1}{11}+\frac{1}{12}+....+\frac{1}{30}>\frac{1}{30}+\frac{1}{30}+...+\frac{1}{30}=\frac{20}{30}=\frac{2}{3}\)

\(\frac{1}{31}+\frac{1}{32}+...+\frac{1}{60}>\frac{1}{60}+\frac{1}{60}+...+\frac{1}{60}=\frac{20}{60}=\frac{1}{3}\)

\(\Rightarrow A>\frac{2}{3}+\frac{1}{3}=1>\frac{1}{2}\)

Vậy A > 1/2

b, Ta có: \(\frac{1}{50}>\frac{1}{100};\frac{1}{51}>\frac{1}{100};........;\frac{1}{99}>\frac{1}{100}\)

\(\Rightarrow B>\frac{1}{100}+\frac{1}{100}+...+\frac{1}{100}=\frac{50}{100}=\frac{1}{2}\)

Vậy B > 1/2

c, Ta có: \(C=\frac{1}{10}+\frac{1}{11}+\frac{1}{12}+...+\frac{1}{100}=\frac{1}{10}+\left(\frac{1}{11}+\frac{1}{12}+...+\frac{1}{100}\right)\)

Nhận xét: \(\frac{1}{11}+\frac{1}{12}+...+\frac{1}{100}>\frac{1}{100}+\frac{1}{100}+...+\frac{1}{100}=\frac{90}{100}=\frac{9}{10}\)

\(\Rightarrow C>\frac{1}{10}+\frac{9}{10}=\frac{10}{10}=1\)

Vậy C > 1

Tớ đồng ý,bạn làm đúng rồi .......

45854

212122512122

1

1

1

1123

4564

454

3546434

Đề sai tại vì:

Ta thấy từ: \(\frac{1}{51}+\frac{1}{52}+\frac{1}{53}+...+\frac{1}{99}\) mỗi số hạng đều lớn hơn \(\frac{1}{100}\)

Mà tổng trên có : ( 100 - 51 ) + 1 = 50 ( số hạng )

Nên:

\(\frac{1}{51}+\frac{1}{52}+\frac{1}{53}+...+\frac{1}{99}+\frac{1}{100}>\frac{1}{100}.50=\frac{50}{100}=\frac{1}{2}\)

Vậy : \(A>\frac{1}{2}\)

\(\Rightarrow3B=3^2+3^3+3^4+...+3^{101}\\ \Rightarrow3B-B=3^2+3^3+...+3^{101}-3-3^2-3^3-...-3^{100}\\ \Rightarrow2B=3^{101}-3\\ \Rightarrow B=\dfrac{3^{101}-3}{2}\)

B = 3¹ + 3² + 3³ + .... + 3⁹⁹ + 3¹⁰⁰

3B = 3(3¹ + 3² + 3³ + ..... + 3⁹⁹ + 3¹⁰⁰)

3B = 3² + 3³ + 3⁴ +...... + 3¹⁰⁰ + 3¹⁰¹

3B - B = 2B = (3² + 3³ + 3⁴ + .... + 3¹⁰⁰ + 3¹⁰¹) - ( 3¹ + 3² + 3³ + .... + 3¹⁰⁰)

2B = (3² - 3²) + (3³ - 3³) +.....+ ( 3¹⁰⁰ - 3¹⁰⁰) + ( 3¹⁰¹ - 1)

2B = 0 + 0 + 0 + ..... +0 + 3¹⁰¹ - 1

2B = 3¹⁰¹ - 1

B = (3¹⁰¹ - 1)  : 2