\(\frac{1}{1.1}+\frac{1}{1.3}+\frac{1}{3.2}+...+\frac{1}{49.25}\) 

\(=\frac{2}{2}.\left(\frac{1}{1.1}+\frac{1}{1.3}+\frac{1}{3.2}+...+\frac{1}{49.25}\right)\)

\(=\frac{2}{1.1.2}+\frac{2}{1.3.2}+\frac{2}{3.2.2}+...+\frac{2}{49.25.2}\)

\(=\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{49.50}\)

\(=2.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\right)\)

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

\(=2.\left(1-\frac{1}{50}\right)\)

\(=2.\frac{49}{50}\)

\(=\frac{49}{25}\)

Chúc bạn học tốt !!! 

3) Ta có : \(A=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+.....+\frac{2}{99.101}\)

\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+.....+\frac{1}{99}-\frac{1}{101}\)

\(=1-\frac{1}{101}=\frac{100}{101}\)

4)

A = \(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{99.101}\)

A = \(\frac{1}{2}.\left(1-\frac{1}{3}\right)+\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{5}\right)+\frac{1}{2}.\left(\frac{1}{5}-\frac{1}{7}\right)+...+\frac{1}{2}.\left(\frac{1}{99}-\frac{1}{101}\right)\)

A = \(\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\right)\)

A = \(\frac{1}{2}.\left(1-\frac{1}{101}\right)\)

\(A=\frac{1}{2}.\frac{100}{101}\)

A = \(\frac{50}{101}\)

2, đặt tên biểu thức trên là A. Ta có :

\(A=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{10100}\)

\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{100.101}\)

\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{100}-\frac{1}{101}\)

\(A=1-\frac{1}{101}\)

\(A=\frac{100}{101}\)

1) \(\frac{1}{1}.\frac{1}{2}+\frac{1}{2}.\frac{1}{3}+\frac{1}{3}.\frac{1}{4}+\frac{1}{4}.\frac{1}{5}\)

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

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

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

Bài 1:

1.1

a) Ta có: \(A=\left(x+y\right)\left(x-y\right)+x\left(2x-1\right)+y\left(y+1\right)\)

\(=x^2-y^2+2x^2-x+y^2+y\)

\(=3x^2-x+y\)

b) Thay x=1 và y=2018 vào biểu thức \(A=3x^2-x+y\), ta được:

\(A=3\cdot1^2-1+2018\)

\(=2+2018=2020\)

Vậy: Khi x=1 và y=2018 thì A=2020

1.2

a) Ta có: \(2x^2\left(x^2-3x+1\right)\)

\(=2x^2\cdot x^2-2x^2\cdot3x+2x^2\cdot1\)

\(=2x^4-6x^3+2x^2\)

b) Ta có: \(\left(2x-1\right)\left(6x^2+3x-3\right)\)

\(=2x\cdot6x^2+2x\cdot3x-2x\cdot3-6x^2-3x+3\)

\(=12x^3+6x^2-6x-6x^2-3x+3\)

\(=12x^3-9x+3\)

1.3

a) Ta có: \(x^3-2x^2+x\)

\(=x\left(x^2-2x+1\right)\)

\(=x\left(x-1\right)^2\)

b) Ta có: \(x^2-xy-8x+8y\)

\(=x\left(x-y\right)-8\left(x-y\right)\)

\(=\left(x-y\right)\left(x-8\right)\)

1.1

a) A= (x+y).(x-y) + x(2x-1) + y(y+1)

= x²- x.y + x.y - y² + 2x² - x +y² + y = 3x² - x + y

b) Ta có A= 3x² - x + y; thay x=1,y=2018 vào biểu thức:

A= 3.1² - 1+ 2018 = 2020

1.3

a)x³ - 2x² + x = x.( x² - 2x + 1) = x.(x-1)²

b) x² - xy - 8x + 8y = x.(x - y) - 8.(x - y)= (x - y).(x-8).

Xin lỗi nha, tớ không biết làm bài 1.2.

Chúc bạn học tốt!!