\(A=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+....+\dfrac{1}{99\cdot100}\)

\(A=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\)

\(A=1+\left(-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}\right)-\dfrac{1}{100}\)

\(A=1+0-\dfrac{1}{100}\)

\(A=1-\dfrac{1}{100}< 1\)

\(\Rightarrow A< 1\)

\(\dfrac{3}{1\times4}+\dfrac{3}{4\times7}+...+\dfrac{3}{226\times229}\)

\(=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+....+\dfrac{1}{226}-\dfrac{1}{229}\)

\(=1+\left(-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+....+\dfrac{1}{226}\right)-\dfrac{1}{229}\)

\(=1+0-\dfrac{1}{229}\)

\(=1-\dfrac{1}{229}\)

\(=\dfrac{229}{229}-\dfrac{1}{229}\)

\(=\dfrac{229-1}{229}\)

\(=\dfrac{228}{229}\)

\(5x+5^0=3x+3^2\\ 5x+1=3x+9\\ 5x-3x=9-1\\ x\left(5-3\right)=8\\ x2=8\\ x=8\div2\\ x=4\)

\(\left(5-x\right)^{2020}=\left(5-x\right)^{2022}\\ \left(5-x\right)^{2020}-\left(5-x\right)^{2022}=0\\ \left(5-x\right)^{2020}-\left(5-x\right)^{2020}\cdot\left(5-x\right)^2=0\\ \left(5-x\right)^{2020}\cdot\left(1-\left(5-x\right)^2\right)=0\)

\(\Rightarrow Th1:\left(5-x\right)^{2020}=0\\ 5-x=0\\ x=5-0\\ x=5\)   \(\Rightarrow Th2:1-\left(5-x\right)^2=0\\ \left(5-x\right)^2=1-0\\\left(5-x\right)^2=1\\ 5-x=1\\ x=5-1\\ x=4 \)

Vậy \(x\in\left\{5;4\right\}\)

\(A=\dfrac{4}{3}+\dfrac{10}{9}+\dfrac{28}{27}+....+\dfrac{\left(3^{99}+1\right)}{3^{99}}\)

\(A=\dfrac{4}{3}+\dfrac{10}{3^2}+\dfrac{28}{3^3}+...+\dfrac{\left(3^{99}+1\right)}{3^{99}}\)

\(A=\left(1+\dfrac{1}{3}\right)+\left(1+\dfrac{1}{3^2}\right)+\left(1+\dfrac{1}{3^3}\right)+...+\left(1+\dfrac{1}{3^{99}}\right)\)

\(A=\left(1+1+....+1\right)+\left(\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{99}}\right)\)

\(A=99+\left(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{99}}\right)\)

Gọi \(\left(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{99}}\right)\)là T

\(T=\left(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{99}}\right)\)

\(3T=1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{98}}\)

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

\(2T=1-\dfrac{1}{3^{99}}\)

\(T=\left(1-\dfrac{1}{3^{99}}\right):2\)

\(T=\dfrac{1}{2}-\dfrac{1}{3^{99}\cdot2}\)

\(=>A=99+T=99+\dfrac{1}{2}-\dfrac{1}{3^{99}\cdot2}=99,5-\dfrac{1}{3^{99}\cdot2}< 100\)

Vậy A < 100

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)

\(\dfrac{3}{1\cdot4}+\dfrac{4}{4\cdot8}+\dfrac{5}{8\cdot13}+..........+\dfrac{10}{43\cdot53}\)

\(=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{13}+....................+\dfrac{1}{43}-\dfrac{1}{53}\)

\(=1-\dfrac{1}{53}\)

\(=\dfrac{52}{53}\)

C

Số số hạng: ( 2020 - 1) : 1 + 1 = 2020 (số)

Tổng: (2020 +1)*2020 : 2 = 2041210 

(3n−1)⋮n−1

=>(3n−1)−3⋅(n−1)⋮n−1

$\left(3n-1\right)-3\cdot \left(n-1\right)$

$=3n-1-3n+3$

$=\left(3n-3n\right)+\left(-1+3\right)$

$=0+2=2$

$=>2⋮�-1=>2⋮n-1$

$=>(�-1)\in Ư(2)=>\left(n-1\right)\in Ư\left(2\right)$

Ư(2) = {-1;1;-2;2}

Th1

n-1=-1

=> n=0

Th2

n-1=1

=> n=2

Th3

n-1=-2

=> n=-1

Th4

n-1=2

=> n= 3

Vậy n ϵ {3;-1;2;0}

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