Question

 Cho x+y=a ;x.y=b

Tính x^2 + y^2; x^3 + y^3; x^4 + y^4; x^5 + y^5

Answer 1

Ta có: 

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

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

\(x^4+y^4=\left(x^2+y^2\right)^2-2x^2y^2=\left(a^2-2b\right)^2-2b^2\)

\(=a^4-4a^2b+4b^2-2b^2=a^4-4a^2b+2b^2\)

\(x^5+y^5=\left(x+y\right)^5-\left(5x^4y+10x^3y^2+10x^2y^3+5xy^4\right)\)

\(=\left(x+y\right)^5-5xy\left(x^3+y^3\right)-10x^2y^2\left(x+y\right)\)

\(=a^5-5\left(a^3-3ab\right)b-10ab^2\)

\(=a^5-5a^3b+15ab^2-10ab^2\)

\(=a^5-5a^3b+5ab^2\)

Answer 2

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

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

\(x^4+y^4=\left(x^2+y^2\right)^2-2x^2y^2=\left[\left(x+y\right)^2-2xy\right]^2-2x^2y^2=\left(a^2-2b\right)^2-2b^2\)

\(=a^2-4a^2b+2b^2\)

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