ta có : \(x^2+y^2=10\Leftrightarrow\left(x+y\right)^2-2xy=10\Leftrightarrow2^2-2xy=10\)
\(\Leftrightarrow4-2xy=10\Leftrightarrow-2xy=10-4=6\Leftrightarrow xy=\dfrac{6}{-2}=-3\)
ta lại có :
* \(x^3+y^3=\left(x+y\right)^3-3xy\left(x+y\right)=2^3-3.\left(-3\right).2\)
\(=8+18=26\) vậy \(x^3+y^3=26\) khi \(x+y=2;x^2+y^2=10\)
* \(x^4+y^4=\left(x+y\right)\left(x^3+x^2y+xy^2+y^3\right)\)
\(=\left(x+y\right)\left(\left(x^3+y^3\right)+\left(x^2y+xy^2\right)\right)\)
\(=\left(x+y\right)\left(\left(\left(x+y\right)^3-3xy\left(x+y\right)+\left(xy.\left(x+y\right)\right)\right)\right)\)
\(=2\left(\left(\left(2\right)^3-3\left(-3\right)\left(2\right)+\left(-3\right)\left(2\right)\right)\right)\)
\(=2\left(\left(\left(8+18\right)-6\right)\right)=2\left(26-6\right)=2.20=40\)
vậy \(x^4+y^4=40\) khi \(x+y=2;x^2+y^2=10\)
\(x^3+y^3=\left(x+y\right)\left(x^2-xy+y^2\right)\\ =\left(x+y\right)\left(x^2+y^2-xy\right)=2\left(10-xy\right)\)
nữa thì hết biết :v