1/2 N=1/2x3 + 1/3x4 +...+1/9x10 

1/2 N=1/2-1/3+1/3-1/4+...+1/9-1/10

1/2 N=1/2-1/10=2/5

N=2/5:1/2=4/5

b) \(x^3-y^3-3xy\)

\(=\left(x-y\right)\left(x^2+xy+y^2\right)-3xy\)

\(=\left(x-y\right)\left[\left(x+y\right)^2-2xy+xy\right]-3xy\)

\(=\left(x-y\right)\left(1-xy\right)-3xy\)

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

A= \(\frac{1+3+5+...+19}{21+23+25+...+39}=\frac{20.5}{60.5}\frac{1}{3}\)

Đặt A= \(\dfrac{1}{10}+\dfrac{1}{15}+\dfrac{1}{21}+......+\dfrac{1}{4950}\)

A.\(\dfrac{1}{2}=\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+......+\dfrac{1}{9900}\)

A.\(\dfrac{1}{2}=\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+......+\dfrac{1}{99.100}\)

A.\(\dfrac{1}{2}=\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+.......+\dfrac{1}{99}\)\(-\dfrac{1}{100}\)

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

A.\(\dfrac{1}{2}=\dfrac{6}{25}\)

A=\(\dfrac{6}{25}:\dfrac{1}{2}\)

A=\(\dfrac{12}{25}\)

Mình giải hơi muộn nhưng tick cho mình với nhé!

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

\(B=1+\left(2-3-4+5\right)+\left(6-7-8+9\right)+...+\left(302-303-304+305\right)+306\)

Các giá trị trong ( ) đều bằng 0 nên:

B = 1 + 306 = 307.