966 Вправа 966Алгебра 8 клас Бевз 2025 · § 19 · стор. 183

966. Розв’яжи рівняння, що зводиться до квадратного. а) 4z(z − 1) = 15; б) 3y(3y − 4) = 5; в) x(7 − x) = 5x − 8; г) 3x(5x + 3) = 2x(6x + 5) + 2; ґ) (2x − 8)(3x + 1) = (4x − 12)(x − 2) + 8.

а) ${\displaystyle 4z(z-1)=}$${\displaystyle 15}$;

${\displaystyle 4z^2-4z=}$${\displaystyle 15}$;

${\displaystyle 4z^2-4z-15=}$${\displaystyle 0}$;

${\displaystyle a=}$${\displaystyle 4,b=}$${\displaystyle -4,c=}$${\displaystyle -15}$;

${\displaystyle D=}$${\displaystyle b^2-4ac=}$${\displaystyle {(-4)}^2-4\cdot 4\cdot (-15)=}$${\displaystyle 16+240=}$${\displaystyle 256}$;

${\displaystyle D>0}$, два корені:

${\displaystyle z_{1,2}=}$${\displaystyle \frac{-b\pm \sqrt{D}}{2a}=}$${\displaystyle \frac{4\pm \sqrt{256}}{8}=}$${\displaystyle \frac{4\pm 16}{8}}$;

${\displaystyle z_1=}$${\displaystyle \frac{4+16}{8}=}$${\displaystyle 2,5}$;

${\displaystyle z_2=}$${\displaystyle \frac{4-16}{8}=}$${\displaystyle -1,5}$;

б) ${\displaystyle 3y(3y-4)=}$${\displaystyle 5}$;

${\displaystyle 9y^2-12y=}$${\displaystyle 5}$;

${\displaystyle 9y^2-12y-5=}$${\displaystyle 0}$;

${\displaystyle a=}$${\displaystyle 9,b=}$${\displaystyle -12,c=}$${\displaystyle -5}$;

${\displaystyle D=}$${\displaystyle b^2-4ac=}$${\displaystyle {(-12)}^2-4\cdot 9\cdot (-5)=}$${\displaystyle 144+180=}$${\displaystyle 324}$;

${\displaystyle D>0}$, два корені:

${\displaystyle y_{1,2}=}$${\displaystyle \frac{-b\pm \sqrt{D}}{2a}=}$${\displaystyle \frac{12\pm \sqrt{324}}{18}=}$${\displaystyle \frac{12\pm 18}{18}}$;

${\displaystyle y_1=}$${\displaystyle \frac{12+18}{18}=}$${\displaystyle \frac{5}{3}}$;

${\displaystyle y_2=}$${\displaystyle \frac{12-18}{18}=}$${\displaystyle -\frac{1}{3}}$;

в) ${\displaystyle x(7-x)=}$${\displaystyle 5x-8}$;

${\displaystyle 7x-x^2=}$${\displaystyle 5x-8}$;

${\displaystyle 7x-x^2-5x+8=}$${\displaystyle 0}$;

${\displaystyle -x^2+2x+8=}$${\displaystyle 0\mid \cdot \left(-1\right)}$;

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

${\displaystyle a=}$${\displaystyle 1,b=}$${\displaystyle -2,c=}$${\displaystyle -8}$;

${\displaystyle D=}$${\displaystyle b^2-4ac=}$${\displaystyle {(-2)}^2-4\cdot 1\cdot (-8)=}$${\displaystyle 4+32=}$${\displaystyle 36}$;

${\displaystyle D>0}$, два корені:

${\displaystyle x_{1,2}=}$${\displaystyle \frac{-b\pm \sqrt{D}}{2a}=}$${\displaystyle \frac{2\pm \sqrt{36}}{2}=}$${\displaystyle \frac{2\pm 6}{2}}$;

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

${\displaystyle x_2=}$${\displaystyle \frac{2-6}{2}=}$${\displaystyle -2}$;

г) ${\displaystyle 3x(5x+3)=}$${\displaystyle 2x(6x+5)+2}$;

${\displaystyle 15x^2+9x=}$${\displaystyle 12x^2+10x+2}$;

${\displaystyle 15x^2+9x-12x^2-10x-2=}$${\displaystyle 0}$;

${\displaystyle 3x^2-x-2=}$${\displaystyle 0}$;

${\displaystyle a=}$${\displaystyle 3,b=}$${\displaystyle -1,c=}$${\displaystyle -2}$;

${\displaystyle D=}$${\displaystyle b^2-4ac=}$${\displaystyle {(-1)}^2-4\cdot 3\cdot (-2)=}$${\displaystyle 1+24=}$${\displaystyle 25}$;

${\displaystyle D>0}$, два корені:

${\displaystyle x_{1,2}=}$${\displaystyle \frac{-b\pm \sqrt{D}}{2a}=}$${\displaystyle \frac{1\pm \sqrt{25}}{6}=}$${\displaystyle \frac{1\pm 5}{6}}$;

${\displaystyle x_1=}$${\displaystyle \frac{1+5}{6}=}$${\displaystyle 1}$;

${\displaystyle x_2=}$${\displaystyle \frac{1-5}{6}=}$${\displaystyle -\frac{2}{3}}$;

ґ) ${\displaystyle (2x-8)(3x+1)=}$${\displaystyle (4x-12)(x-2)+8}$;

${\displaystyle 6x^2+2x-24x-8=}$${\displaystyle 4x^2-8x-12x+24+8}$;

${\displaystyle 6x^2-22x-8=}$${\displaystyle 4x^2-20x+32}$;

${\displaystyle 2x^2-2x-40=}$${\displaystyle 0\mid :2}$;

${\displaystyle x^2-x-20=}$${\displaystyle 0}$;

${\displaystyle a=}$${\displaystyle 1,b=}$${\displaystyle -1,c=}$${\displaystyle -20}$;

${\displaystyle D=}$${\displaystyle b^2-4ac=}$${\displaystyle {(-1)}^2-4\cdot 1\cdot (-20)=}$${\displaystyle 1+80=}$${\displaystyle 81}$;

${\displaystyle D>0}$, два корені:

${\displaystyle x_{1,2}=}$${\displaystyle \frac{-b\pm \sqrt{D}}{2a}=}$${\displaystyle \frac{1\pm \sqrt{81}}{2}=}$${\displaystyle \frac{1\pm 9}{2}}$;

${\displaystyle x_1=}$${\displaystyle \frac{1+9}{2}=}$${\displaystyle 5}$;

${\displaystyle x_2=}$${\displaystyle \frac{1-9}{2}=}$${\displaystyle -4}$.