Question

Помогите пожалуйста решить

Answer 1

Ответ:

$\displaystyle \left \{ {{u_1=-3} \atop {v_1=3.5}} \right. \quad \left \{ {{u_2=-2} \atop {v_2=3}}$

Пошаговое объяснение:

$\displaystyle \left \{ {{u+2v=4\hfill} \atop {u^2+uv-v=5}} \right. \left \{ {{u=4-2v\hfill} \atop {(4-2v)^2+v(4-2v)-v=-5}} \right. \\\\\\16-16v+4v^2+4v-2v^2-v+5=0\\\\2v^2-13v+21=0\\\\D=b^2-4ac = 13^2-4*2*21=169-168=1\\\\v_1=\frac{-b+\sqrt{D} }{2a} =\frac{13+1}{4} =3.5\\\\\\v_2=\frac{-b-\sqrt{D} }{2a} =\frac{13-1}{4} =3\\\\\\u_1=4-2v_1=4-7=-3\\\\\\u_2=4-2v_2=4-6=-2\\\\\\\left \{ {{u_1=-3} \atop {v_1=3.5}} \right. \quad \left \{ {{u_2=-2} \atop {v_2=3}} \right.$