Question

proszę o pomoc matematyka​

Answer 1

Odpowiedź:

Szczegółowe wyjaśnienie:

1)

[tex]\left \{ {{x^{2} -y^{2} =13} \atop {x-5y-13=0}} \right. \\\\\left \{ {{x^{2} -y^{2} =13} \atop {x=5y+13}} \right. \\\\\left \{ {{(5y+13)^{2} -y^{2} =13} \atop {x=5y+13}} \right. \\\\\left \{ {{25y^{2}+130y+169 -y^{2}-13 =0} \atop {x=5y+13}} \right. \\\\\left \{ {{24y^{2}+130y+156=0} \atop {x=5y+13}} \right. \\\\[/tex]

Δ=130*130-4*24*156=1924

[tex]y_{1} =\frac{-130-2\sqrt{481}}{48} =\frac{-65-\sqrt{481} }{24} \\\\y_{2} =\frac{-130+2\sqrt{481}}{48} =\frac{-65+\sqrt{481} }{24}[/tex]

[tex]x_{1} =5*\frac{-65-\sqrt{481} }{24}+13=\frac{-13-5\sqrt{481} }{24} \\\\x_{2} =5*\frac{-65+\sqrt{481} }{24}+13=\frac{-13+5\sqrt{481} }{24} \\[/tex]

2)

[tex]\left \{ {{y=x+4} \atop {(x-1)^{2}+(y-3)^{2}}=4} \right. \\\\\left \{ {{y=x+4} \atop {(x-1)^{2}+(x+4-3)^{2}}=4} \right. \\\\\left \{ {{y=x+4} \atop {(x-1)^{2}+(x+1)^{2}}=4} \right. \\\\\left \{ {{y=x+4} \atop {x^{2} -2x+1+x^{2} +2x+1=4} \right. \\\\\left \{ {{y=x+4} \atop {2x^{2} +2=4} \right. \\\\\left \{ {{y=x+4} \atop {x^{2} -1=0} \right. \\\\\left \{ {{y=x+4} \atop {(x-1)(x+1)=0} \right. \\[/tex]

[tex]x_{1} =1\\x_{2}=-1[/tex]    [tex]y_{1} =5\\y_{2} =3[/tex]