Question

a) Cho \(x+y=1\) và \(xy=-6\). Tính \(x^2+y^2;x^3+y^3;x^5+y^5\).

b) Cho \(x-y=1\) và \(xy=6\).Tính \(x^2-y^2;x^3-y^3;x^5-y^5\).

Answer 1

a,

\(x^2+y^2=\left(x+y\right)^2-2xy=1^2-2\cdot\left(-6\right)=1-\left(-12\right)=13\)

\(x^3+y^3=\left(x+y\right)\left(x^2-xy+y^2\right)=1\cdot\left[13-\left(-6\right)\right]=19\)

\(x^5+y^5=\left(x+y\right)\left(x^2+y^2\right)^2-\left(2x^3y^2+xy^4+x^4y+2x^2y^3\right)=169-\left[2\left(xy\right)^2\left(x+y\right)+xy\left(x^3+y^3\right)\right]=169-\left[2\cdot36\cdot1-6\cdot19\right]=211\)

Answer 2

b,

\(x^2+y^2=\left(x-y\right)^2+2xy=1+12=13\)

\(x^3-y^3=\left(x-y\right)\left(x^2+xy+y^2\right)=1\cdot\left(13+6\right)=19\)