1395 Вправа 1395Алгебра 8 клас Бевз 2025 · Повторення · стор. 264

1395. Розв’яжи систему рівнянь. а) 2x² − y = 2, x − y = 1; б) x + y = 8, x² + y² = 40; в) x² − y = 14, y + 2 = x.

Спосіб підстановки: з лінійного рівняння виражаємо ${\displaystyle y}$ і підставляємо в друге.

а) ${\displaystyle y=}$${\displaystyle x-1}$:

${\displaystyle 2x^2-(x-1)=}$${\displaystyle 2}$;

${\displaystyle 2x^2-x+1-2=}$${\displaystyle 0}$;

${\displaystyle 2x^2-x-1=}$${\displaystyle 0}$;

${\displaystyle a=}$${\displaystyle 2,b=}$${\displaystyle -1,c=}$${\displaystyle -1}$;

${\displaystyle D=}$${\displaystyle b^2-4ac=}$${\displaystyle {(-1)}^2-4\cdot 2\cdot (-1)=}$${\displaystyle 1+8=}$${\displaystyle 9}$;

${\displaystyle D>0}$, два корені:

${\displaystyle x_{1,2}=}$${\displaystyle \frac{-b\pm \sqrt{D}}{2a}=}$${\displaystyle \frac{1\pm \sqrt{9}}{4}=}$${\displaystyle \frac{1\pm 3}{4}}$;

${\displaystyle x_1=}$${\displaystyle \frac{1+3}{4}=}$${\displaystyle 1}$;

${\displaystyle x_2=}$${\displaystyle \frac{1-3}{4}=}$${\displaystyle -0,5}$;

${\displaystyle y=}$${\displaystyle 1-1=}$${\displaystyle 0}$;

${\displaystyle y=}$${\displaystyle -0,5-1=}$${\displaystyle -1,5}$;

розв’язки: ${\displaystyle \left(1;0\right)}$, ${\displaystyle \left(-0,5;-1,5\right)}$;

б) ${\displaystyle y=}$${\displaystyle 8-x}$:

${\displaystyle x^2+{(8-x)}^2=}$${\displaystyle 40}$;

${\displaystyle 2x^2-16x+64-40=}$${\displaystyle 0}$;

${\displaystyle 2x^2-16x+24=}$${\displaystyle 0\mid :2}$;

${\displaystyle x^2-8x+12=}$${\displaystyle 0}$;

${\displaystyle a=}$${\displaystyle 1,b=}$${\displaystyle -8,c=}$${\displaystyle 12}$;

${\displaystyle D=}$${\displaystyle b^2-4ac=}$${\displaystyle {(-8)}^2-4\cdot 1\cdot 12=}$${\displaystyle 64-48=}$${\displaystyle 16}$;

${\displaystyle D>0}$, два корені:

${\displaystyle x_{1,2}=}$${\displaystyle \frac{-b\pm \sqrt{D}}{2a}=}$${\displaystyle \frac{8\pm \sqrt{16}}{2}=}$${\displaystyle \frac{8\pm 4}{2}}$;

${\displaystyle x_1=}$${\displaystyle \frac{8+4}{2}=}$${\displaystyle 6}$;

${\displaystyle x_2=}$${\displaystyle \frac{8-4}{2}=}$${\displaystyle 2}$;

розв’язки: ${\displaystyle \left(6;2\right)}$, ${\displaystyle \left(2;6\right)}$;

в) ${\displaystyle y=}$${\displaystyle x-2}$:

${\displaystyle x^2-(x-2)=}$${\displaystyle 14}$;

${\displaystyle x^2-x+2-14=}$${\displaystyle 0}$;

${\displaystyle x^2-x-12=}$${\displaystyle 0}$;

${\displaystyle a=}$${\displaystyle 1,b=}$${\displaystyle -1,c=}$${\displaystyle -12}$;

${\displaystyle D=}$${\displaystyle b^2-4ac=}$${\displaystyle {(-1)}^2-4\cdot 1\cdot (-12)=}$${\displaystyle 1+48=}$${\displaystyle 49}$;

${\displaystyle D>0}$, два корені:

${\displaystyle x_{1,2}=}$${\displaystyle \frac{-b\pm \sqrt{D}}{2a}=}$${\displaystyle \frac{1\pm \sqrt{49}}{2}=}$${\displaystyle \frac{1\pm 7}{2}}$;

${\displaystyle x_1=}$${\displaystyle \frac{1+7}{2}=}$${\displaystyle 4}$;

${\displaystyle x_2=}$${\displaystyle \frac{1-7}{2}=}$${\displaystyle -3}$;

${\displaystyle y=}$${\displaystyle 4-2=}$${\displaystyle 2}$;

${\displaystyle y=}$${\displaystyle -3-2=}$${\displaystyle -5}$;

розв’язки: ${\displaystyle \left(4;2\right)}$, ${\displaystyle \left(-3;-5\right)}$.