Question

реши систему уравнений методом алгебраического сложения
{t² + q² =13
{t² -q² = 5
t¹ =
q² =

t² =
q² =

t³=
q³=

t⁴ =
q⁴ =

Answer 1

$\left \{ {{t^{2}+q^{2}=13} \atop {t^{2}-q^{2}=5}} \right. \\\\t^{2}+q^{2}+t^{2}-q^{2}=13+5,\\2*t^{2}=18,\\t^{2}=9,\\t_{1} =3, \\t_{2} =-3.$

Тогда (±3)²+q²=13, q²+9=13, q²=4, q=±2.

$t^{1} =$ ±3,

$q^{1} =$ ±2.

$t^{2} =9,\\q^{2}=4.$

$t^{3} =$ ±27,

$q^{3} =$ ±8.

$t^{4} =81,\\q^{4}=16.$