Question

cho x + y=a ; xy=b. Tính giá trị của biểu thức sau theo a,b : 

x⁵ + y⁵

Answer 1

Ta có: \(x^5+y^5=\left(x^2+y^2\right)\left(x^3+y^3\right)-x^2y^3-x^3y^2\) 

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

\(=\left(a^2-2b\right)\left(a^3-3ab\right)-ab^2\) 

 \(=a^5-5a^3b+5ab^2\) 

\(=a^5-5ab\left(a^2-b\right)\)

Answer 2

\(x^5+y^5=\left(x^2+y^2\right)\left(x^3+y^3\right)-x^2y^3-x^3y^2\)

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

\(=\left(a^2-2b\right)\left(a^3-3ab\right)-ab^2\)

\(=a^5-5ab\left(a^2-b\right)\)