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 $x+y=2;x^2+y^2=10$