b) \(\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{2021.2023}\)

\(=\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{2021}-\dfrac{1}{2023}\)

\(=\dfrac{1}{1}-\dfrac{1}{2023}\)

\(=\dfrac{2022}{2023}\)

\(b)\)\(A=\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{2021.2023}\)

\(2A=\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{2021.2023}\)

\(2A=\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{2021}-\dfrac{1}{2023}\)

\(2A=\dfrac{1}{1}-\dfrac{1}{2023}\)

\(2A=\dfrac{2022}{2023}\)

\(A=\dfrac{2022}{2023}:2\)

\(A=\dfrac{1011}{2023}\)

tính tổng 

\(=\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{2003.2005}\)

\(=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2003}-\frac{1}{2005}\right)\)

\(=\frac{1}{2}.\left(1-\frac{1}{2005}\right)\)

\(=\frac{1}{2}.\frac{2004}{2005}\)

\(=\frac{1002}{2005}\)

yêu Sáng

a.Từ giả thiết: 
x+y=1. 
=> (x+y)^3=1^3=1 
=> x^3 +3x^2.y+3x.y^2+y^3=1(HĐT) 
=> x^3+y^3+3xy(x+y)=1 
=> x^3+y^3+3xy.1=1 
<=> x^3+y^3+3xy=1

b.x³-y³-3xy=x³-y³-3xy.1

Mà x-y=1 nên

x³-y³-3xy=x³-y³-3xy(x-y)

x³-y³-3x²y+3xy² 

=(x-y)³=1³=1

Đặt :

\(A=\dfrac{1}{1.3}+\dfrac{1}{3.5}+....+\dfrac{1}{2003.2005}\)

\(\Leftrightarrow2A=\dfrac{2}{1.3}+\dfrac{2}{3.5}+.......+\dfrac{2}{2003.2005}\)

\(\Leftrightarrow2A=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+.......+\dfrac{1}{2003}-\dfrac{1}{2005}\)

\(\Leftrightarrow2A=1-\dfrac{1}{2005}\)

\(\Leftrightarrow2A=\dfrac{2004}{2005}\)

\(\Leftrightarrow A=\dfrac{1002}{2005}\)

\(M=\dfrac{1}{1.3}+\dfrac{1}{3.5}+...+\dfrac{1}{2003.2005}\)

\(=\dfrac{1}{2}\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+...+\dfrac{2}{2003.2005}\right)\)

\(=\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{2003}-\dfrac{1}{2005}\right)\)

\(=\dfrac{1}{2}\left(1-\dfrac{1}{2005}\right)\)

\(=\dfrac{1}{2}.\dfrac{2004}{2005}=\dfrac{1002}{2005}\)

2M= 1/1.3+1/3.5+1/5.7+...+1/2003.2005

2M= 1/1-1/3+1/3-1/5+...+1/2003-1/2005

2M= 1/1-1/2005

2M= 2004/2005

M= 2004/2005:2

M=1002/2005

\(M=\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{2003.2005}\)

= \(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{2003}-\dfrac{1}{2005}\)

= \(1-\dfrac{1}{2005}\)

= \(\dfrac{2004}{2005}\)

Đề sai rồi bạn

Sửa đề: f(x) = x² - 4x + 3

a) f(0) = 0 - 4.0 + 3 = 3

f(1) = 1 - 4.1 + 3 = 0

f(3) = 9 - 4.3 + 3 = 0

b) x = 1 và x = 3 là nghiệm của đa thức f(x) vì f(1) = 0 và f(3) = 0

Đặt biểu thức là A

\(2A=\dfrac{3-1}{1.3}+\dfrac{5-3}{3.5}+\dfrac{7-5}{5.7}+...+\dfrac{2005-2003}{2003.2005}=\)

\(=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{2003}-\dfrac{1}{2005}=1-\dfrac{1}{2005}=\dfrac{2004}{2005}\)

\(\Rightarrow A=\dfrac{2004}{2005}:2=\dfrac{1002}{2005}\)

Gọi tổng trên là A. Ta có

2A=\(\dfrac{2}{1.3}+\dfrac{2}{3.5}+...+\dfrac{2}{2003.2005}\)

2A=\(\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{2003}-\dfrac{1}{2005}\)

2A=\(\dfrac{1}{1}-\dfrac{1}{2005}=\dfrac{2005}{2005}-\dfrac{1}{2005}=\dfrac{2004}{2005}\)

⇒ A= \(\dfrac{2004}{2005}:2=\dfrac{2004}{2005}.\dfrac{1}{2}=\dfrac{1002}{2005}\)

Vậy tổng trên bằng \(\dfrac{1002}{2005}\)

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