1) Cho x+y=2 và x^2+y^2=10. Tính x^3+y^3. Giải
(x+y)^2=x^2+y^2+2xy => xy= -3
x^3+y^3=(x+y)^3-3xy(x+y) = 26
2) Ta có: x^3+y^3 = (x+y)(x^2-xy+y^2) (1)
(x+y)^2=a^2
=> x^2 +2xy +y^2=a^2
=> b+2xy=a^2
=> xy=\(\frac{a^2-b}{2}\)
Thay (1) vào đó ta có:
x^3+y^3= (x+y)(x^2-xy+y^2) = a(b-\(\frac{a^2-b}{2}\)) = \(a\left(\frac{2b-a^2+b}{2}\right)=a.\frac{3b-a^2}{2}\)
\(x^3+y^3=\left(x+y\right)\left(x^2-xy+y^2\right)=2\left(10-xy\right)\)
Ta có: \(x^2+y^2=\left(x+y\right)^2-2xy=2^2-2xy=4-2xy=10\Rightarrow2xy=-6\Rightarrow xy=-3\)
Vậy: \(x^3+y^3=2\left(10+3\right)=2.13=26\)
\(x^3+y^3=\left(x+y\right)\left(x^2-xy+y^2\right)=a\left(b-xy\right)\)
