Решите систему уравнений:
а)у=2х-5
х^2+у^2=25
б)у=х^2-4х+1
2х+у=4

а)y=2x-5x^2+y^2=25x^2+(2x-5)^2=25x^2+4x^2-20x+25 =255x^2-20x = 05x^2=20xx^2=25x=5y=5б)y=x^2-4x+1  2x+y=4    2x+x^2-4x+1 = 4    x^2-2x-3=0    x1=1    x2=3    y1=2    y2=-2Вроде как то так)))

a) y=2x–5x^2+y^2=25x^2+(2x–5)^2=25x^2+4x^2–20x+25–25=05x^2–20x=05x(x–4)=05x=0; x–4=0x1=0; x2=4y1=–5; y2=3б) у=х^2–4х+12х+у=4; у=4–2х2х+х^2–4х+1–4=0х^2–2х–3=0Д=/4–4•1•(-3)=/16=4х1=(2+4)/2=3; х2=(2–4)/2=–1у1=4–2•3=–2; у2=4–2•(-1)=6