\(A=\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{49.50}\)

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

\(A=\dfrac{1}{1}-\dfrac{1}{50}\)

\(A=\dfrac{49}{50}\)

A = 49/50

A = 1/1.2 +1/2.3 +1/3.4 +...+1/49.50    
A = 1 +1/2 -1/2+1/3-1/3+1/4-...-1/49 +1/50    

A = 1 - 1/50   
A=49/50

a: \(=\dfrac{1}{1\cdot2}-\dfrac{1}{2\cdot3}+\dfrac{1}{2\cdot3}-\dfrac{1}{3\cdot4}+...+\dfrac{1}{18\cdot19}-\dfrac{1}{19\cdot20}\)

=1/2-1/380

=179/380

b: \(=\dfrac{1}{1\cdot3}-\dfrac{1}{3\cdot5}+\dfrac{1}{3\cdot5}-\dfrac{1}{5\cdot7}+...+\dfrac{1}{21\cdot23}-\dfrac{1}{23\cdot25}\)

\(=\dfrac{1}{3}-\dfrac{1}{575}=\dfrac{572}{1725}\)

c: \(=1+\dfrac{1}{2}-\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{19}+\dfrac{1}{20}-\dfrac{1}{20}-\dfrac{1}{21}\)

=1-1/21

=20/21

d: \(=\left(1-\dfrac{1}{9}\right)\left(1-\dfrac{1}{16}\right)\cdot...\cdot\left(1-\dfrac{1}{121}\right)\)

\(=\dfrac{2}{3}\cdot\dfrac{3}{4}\cdot...\cdot\dfrac{10}{11}\cdot\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot...\cdot\dfrac{12}{11}\)

\(=\dfrac{2}{11}\cdot\dfrac{12}{2}=\dfrac{12}{11}\)

\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}\)

=\(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}\)

=\(1-\dfrac{1}{5}\)

=\(\dfrac{4}{5}\)

1/1 - 1/2+ 1/2 - 1/3 + 1/3 - 1/4+ 1/4 - 1/5

 = 1/1 - 1/5

= 4/5

\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}=1-\dfrac{1}{5}=\dfrac{4}{5}\)

a) A = \(\dfrac{1^2}{1.2}.\dfrac{2^2}{2.3}.\dfrac{3^2}{3.4}.\dfrac{4^2}{4.5}\)

A = \(\dfrac{1.1}{1.2}.\dfrac{2.2}{2.3}.\dfrac{3.3}{3.4}.\dfrac{4.4}{4.5}\)

A = \(\dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}.\dfrac{4}{5}\)= \(\dfrac{1}{5}\)

b) B = \(\dfrac{2^2}{1.3}.\dfrac{3^2}{2.4}.\dfrac{4^2}{3.5}.\dfrac{5^2}{4.6}\)

B = \(\dfrac{2.3.4.5}{1.2.3.4}.\dfrac{2.3.4.5}{3.4.5.6}\)= \(\dfrac{5}{3}\)

Đầu tiên thì nhắc lại cái hằng đẳng thức cho bạn nào chưa học này: (a-b)²=a²-2ab+b²<=>a²+b²=(a-b)²+2ab

\(S=\dfrac{\left(1^2+2^2\right)}{1.2}+\dfrac{\left(2^2+3^2\right)}{2.3}+...+\dfrac{\left(9^2+10^2\right)}{9.10}\)
\(=\dfrac{\left(\left(1-2\right)^2+2.1.2\right)}{1.2}+\dfrac{\left(\left(2-3\right)^2+2.2.3\right)}{2.3}+...+\dfrac{\left(\left(9-10\right)^2+2.9.10\right)}{9.10}\)
\(=\dfrac{\left(\left(-1\right)^2\right)}{1.2+2}+\dfrac{\left(\left(-1\right)^2\right)}{2.3+2}+...+\dfrac{\left(\left(-1^2\right)\right)}{9.10+2}\)
\(=\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{9.10}+2.9\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{9}-\dfrac{1}{10}+18\)
\(=1-\dfrac{1}{10}+18\)
\(=18,9=\dfrac{189}{10}.\)

~ K chắc là đúng đâu ~

a) \(A=\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{100}}\)

\(2A=2\cdot\left(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{100}}\right)\)

\(2A=1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{101}}\)

\(2A-A=1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{99}}-\dfrac{1}{2}-\dfrac{1}{2^2}-...-\dfrac{1}{2^{100}}\)

\(A=1-\dfrac{1}{2^{100}}\)

b) \(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{2023\cdot2024}\)

\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{2023}-\dfrac{1}{2024}\)

\(=1-\dfrac{1}{2024}\)

\(=\dfrac{2024}{2024}-\dfrac{1}{2024}\)

\(=\dfrac{2023}{2024}\)

cứu 

okee laaa

1)\(\dfrac{1}{2\cdot5}+\dfrac{1}{5\cdot8}+...+\dfrac{1}{x\left(x+3\right)}=\dfrac{11}{70}\)

\(\left(\dfrac{3}{2\cdot5}+\dfrac{3}{5\cdot8}+...+\dfrac{3}{x\left(x+3\right)}\right):3=\dfrac{11}{70}\)

\(\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+.....+\dfrac{1}{x}-\dfrac{1}{x+3}\right)=\dfrac{11}{70}\cdot3\)

\(\dfrac{1}{2}-\dfrac{1}{x+3}=\dfrac{33}{70}\)

\(\dfrac{1}{x+3}=\dfrac{1}{2}-\dfrac{33}{70}\)

\(\dfrac{1}{x+3}=\dfrac{2}{70}\)

\(\dfrac{1}{x+3}=\dfrac{1}{35}\)

\(x+3=35\\ x=35-3\\ x=32\)

2) Số góc đc tạo thành từ 2023 tia chung gốc là:\(\dfrac{2023\cdot2022}{2}=2045253\) (góc)

Bài 1 thì bạn Ánh làm đúng rồi

Bài 2 thì giải chi tiết như này em nhé:

Cứ 1 tia tạo với 2023 - 1 tia còn lại là 2023 - 1 góc

Với 2023 tia thì tạo được số góc là:  (2023 - 1)\(\times\) 2023 góc

Theo cách tính trên thì mỗi góc đã được tính hai lần

Vậy số góc tạo được là: (2023-1)\(\times\) 2023: 2 = 2045253 (góc)

Kết luận: ...

1)12⋅5+15⋅8+...+1�(�+3)=11702⋅51​+5⋅81​+...+x(x+3)1​=7011​

(32⋅5+35⋅8+...+3�(�+3)):3=1170(2⋅53​+5⋅83​+...+x(x+3)3​):3=7011​

(12−15+15−18+.....+1�−1�+3)=1170⋅3(21​−51​+51​−81​+.....+x1​−x+31​)=7011​⋅3

12−1�+3=337021​−x+31​=7033​

1�+3=12−3370x+31​=21​−7033​

1�+3=270x+31​=702​

1�+3=135x+3

Câu 2: 

a: =>-11/12x=-1/6-3/4=-2/12-9/12=-11/12

=>x=1

b: =>x-42=57-x-50=7-x

=>2x=49

hay x=49/2

d: =>x+1=3 hoặc x+1=-3

=>x=2 hoặc x=-4

e: =>2x+3=5 hoặc 2x+3=-5

=>2x=2 hoặc 2x=-8

=>x=1 hoặc x=-4