\(\text{Có 3 trường hợp có thể xảy ra:}\)

\(A=B\)

\(A< B\)
\(A>B\)

mik cần giải mà 

\(A=\left(\frac{1}{31}+\frac{1}{32}+...+\frac{1}{40}\right)+\left(\frac{1}{41}+\frac{1}{42}+...+\frac{1}{50}\right)+\left(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{60}\right)\)

\(Mà:\frac{1}{31}+\frac{1}{32}+...+\frac{1}{40}>\frac{1}{40}.10=\frac{1}{4}\left(\text{10 số hạng}\right)\)

\(\text{Tương tự}:\frac{1}{41}+\frac{1}{42}+...+\frac{1}{50}>\frac{1}{5}\)

\(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{60}>\frac{1}{6}\)

\(\Rightarrow A>\frac{1}{4}+\frac{1}{5}+\frac{1}{6}\)

\(\Rightarrow A>\frac{37}{60}\)

\(Mà\)\(\frac{1}{4}+\frac{1}{5}+\frac{1}{6}>\frac{3}{5}\)

\(\Rightarrow A>\frac{3}{5}\)

\(A=\left(\frac{1}{31}+\frac{1}{32}+...+\frac{1}{40}\right)+\left(\frac{1}{41}+\frac{1}{42}+...+\frac{1}{50}\right)+\left(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{60}\right)\)

\(\frac{1}{31}+\frac{1}{32}+...+\frac{1}{40}< \frac{1}{31}.10=\frac{10}{30}=\frac{1}{3}\left(\text{10 số hạng}\right)\)

\(\Rightarrow A< \frac{4}{5}\)

\(\Rightarrow\frac{3}{5}< A< \frac{4}{5}\)

\(\text{Mik chỉ pít làm z!!!☺}\)

\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2019.2020}\)và 1

gọi

 \(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2019.2020}\)

\(A=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2019}-\frac{1}{2020}\)

\(A=\frac{1}{1}-\frac{1}{2020}=\frac{2019}{2020}\)

VÌ \(\frac{2019}{2020}< 1\Rightarrow A< 1\)

VẬY \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2019.2020}< 1\)

1. a) P = 4 - ( x - 2 )³²

( x - 2 )³² ≥ 0 ∀ x => - ( x - 2 )³² ≤ 0 ∀ x

=> 4 - ( x - 2 )³² ≤ 4 ∀ x

Dấu bằng xảy ra <=> x - 2 = 0 => x = 2

Vậy P_Max = 4 khi x = 2

b) Q = 20 - | 3 - x |

| 3 - x |  ≥ 0 ∀ x => - | 3 - x | ≤ 0 ∀ x

=> 20 - | 3 - x |  ≤ 20 ∀ x

Dấu bằng xảy ra <=> 3 - x = 0 => x = 3

Vậy Q_Max = 20 khi x = 3

c) C = \(\frac{5}{\left(x-3\right)^2+1}\)

Để C có GTLN => ( x - 3 )² + 1 nhỏ nhất dương

=> ( x - 3 )² + 1 = 1

=> ( x - 3 )² = 0

=> x - 3 = 0 

=> x = 3

=> C_Max = \(\frac{5}{\left(3-3\right)^2+1}=\frac{5}{1}=5\)khi x = 3

2. a) \(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{2019\cdot2020}\)

\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2019}-\frac{1}{2020}\)

\(=\frac{1}{1}-\frac{1}{2020}=\frac{2019}{2020}< 1\)

b) \(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^{1000}}\)

\(2A=2\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^{1000}}\right)\)

\(2A=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{999}}\)

\(2A-A=A=\left(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{999}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{1000}}\right)\)

\(A=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{999}}-\frac{1}{2}-\frac{1}{2^2}-\frac{1}{2^3}-...-\frac{1}{2^{1000}}\)

\(A=1-\frac{1}{2^{1000}}< 1\left(đpcm\right)\)

Bài 15 :

a) Đặt \(A=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{2019\cdot2020}\)

\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2019}-\frac{1}{2020}\)

\(A=1-\frac{1}{2020}=\frac{2019}{2020}< \frac{2020}{2020}=1\)

b) Ta có : \(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^{1000}}\)

\(2A=\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^{1001}}\)

\(2A-A=\left(\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^{1001}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{1000}}\right)\)

\(A=\frac{1}{2^{1001}}-\frac{1}{2}\)

Tới đây là so sánh đi nhé

Cái này mình làm hôm qua rồi mà '-'

a) Đặt \(A=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{2019\cdot2020}\)

\(A=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2019}-\frac{1}{2020}\)

\(A=\frac{1}{1}-\frac{1}{2020}=\frac{2019}{2020}\)

\(\Rightarrow A< 1\)

b) \(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^{1000}}\)

\(2A=2\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^{1000}}\right)\)

\(2A=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{999}}\)

\(2A-A=A\)

\(=\left(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{999}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{1000}}\right)\)

\(=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^{999}}-\frac{1}{2}-\frac{1}{2^2}-\frac{1}{2^3}-...-\frac{1}{2^{1000}}\)

\(=1-\frac{1}{2^{1000}}\)

\(\Rightarrow A=1-\frac{1}{2^{1000}}< 1\left(đpcm\right)\)

Bài làm:

a) Ta có: \(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2019.2020}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2019}-\frac{1}{2020}\)

\(=1-\frac{1}{2020}< 1\)

b) \(2A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{999}}\)

\(\Rightarrow2A-A=\left(1+\frac{1}{2}+...+\frac{1}{2^{999}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{1000}}\right)\)

\(\Leftrightarrow A=1-\frac{1}{2^{1000}}< 1\)

A = \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2016.2017}\)

A = \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2016}-\frac{1}{2017}\)

A = \(1-\left(\frac{1}{2}-\frac{1}{2}\right)-\left(\frac{1}{3}-\frac{1}{3}\right)-...-\left(\frac{1}{2016}-\frac{1}{2016}\right)-\frac{1}{2017}\)

A = \(1-0-0-0...-0-\frac{1}{2017}\)

A = \(1-\frac{1}{2017}< 1\)

\(A=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2016.2017}\)

\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2016}-\frac{1}{2017}\)

\(A=1-\frac{1}{2017}=\frac{2016}{2017}< \frac{2017}{2017}=1\)

=> A<1(đpcm)

\(B=\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{59.60}\)

\(B=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{59}-\frac{1}{60}\)

\(B=\left(1+\frac{1}{3}+...+\frac{1}{59}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{60}\right)\)

\(B=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+..+\frac{1}{59}+\frac{1}{60}\right)-2.\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{50}\right)\)

\(B=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{49}+\frac{1}{50}\right)-\left(1+\frac{1}{2}+...+\frac{1}{30}\right)\)

\(B=\frac{1}{31}+\frac{1}{32}+\frac{1}{33}+...+\frac{1}{60}=A\)

A > B nhé tích mk vs

cậu giải giúp tớ cái 

B = 1/1.2 + 1/3.4 + 1/5.6 + ... + 1/59.60

B = 1 - 1/2 + 1/3 - 1/4 + 1/5 - 1/6 + ... + 1/59 - 1/60

B = (1 + 1/3 + 1/5 + ... + 1/59) - (1/2 + 1/4 + 1/6 + ... + 1/60)

B = (1 + 1/2 + 1/3 + 1/4 + 1/5 + 1/6 + ... + 1/59 + 1/60) - 2.(1/2 + 1/4 + 1/6 + ... + 1/60)

B = (1 + 1/2 + 1/3 + 1/4 + 1/5 + 1/6 + ... + 1/59 + 1/60) - (1 + 1/2 + 1/3 + ... + 1/30)

B = 1/31 + 1/32 + 1/33 + ... + 1/60 = A

=> B = A

ta có: Lớn nhất của A là:\(\frac{1}{31}+\frac{1}{31}+...+\frac{1}{31}\)(30 phân số)

         =30/31

  B=1-\(\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{3}+...+\frac{1}{59}-\frac{1}{60}\)\(=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{59}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{60}\right)\)

Bé nhất của của B là :\(\left(1+1+...+1\right)-\left(\frac{1}{60}+\frac{1}{60}+...+\frac{1}{60}\right)\)

                                \(=30-\frac{30}{60}\)

=>B>A

Câu a: Không hỏi nên không trả lời

Câu b:Gọi d là ƯCLN của n và n+1

Ta có: n chia hết cho d

n+1 chia hết cho d

=>(n+1)-n chia hết cho d

=>1 chia hết cho d

=>d=1

Vậy phân số n/n+1 là phân số tối giản

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

=\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)

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

Vì: \(1-\frac{1}{50}\)<\(1\)

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

A>4/5