998 Вправа 998Алгебра 8 клас Бевз 2025 · § 19 · стор. 187

998. Розв’яжи рівняння. а) (x² + x)(x² + x − 7) = 60; б) (2x² − 5x − 4)(2x² − 5x) = 21; в) x(x + 1)(x + 2)(x + 3) = 24; г) (x − 1)²(x² − 2x) = 12; ґ) (x² + 2x + 1)(x² + 2x + 4) = 10; д) (x − 2)(x + 1)(x + 2)(x + 5) + 36 = 0.

а) ${\displaystyle (x^2+x)(x^2+x-7)=}$${\displaystyle 60}$;

заміна ${\displaystyle x^2+x=}$${\displaystyle t}$:

${\displaystyle t(t-7)=}$${\displaystyle 60}$;

${\displaystyle t^2-7t-60=}$${\displaystyle 0}$;

${\displaystyle a=}$${\displaystyle 1,b=}$${\displaystyle -7,c=}$${\displaystyle -60}$;

${\displaystyle D=}$${\displaystyle b^2-4ac=}$${\displaystyle {(-7)}^2-4\cdot 1\cdot (-60)=}$${\displaystyle 49+240=}$${\displaystyle 289}$;

${\displaystyle D>0}$, два корені:

${\displaystyle t_{1,2}=}$${\displaystyle \frac{-b\pm \sqrt{D}}{2a}=}$${\displaystyle \frac{7\pm \sqrt{289}}{2}=}$${\displaystyle \frac{7\pm 17}{2}}$;

${\displaystyle t_1=}$${\displaystyle \frac{7+17}{2}=}$${\displaystyle 12}$;

${\displaystyle t_2=}$${\displaystyle \frac{7-17}{2}=}$${\displaystyle -5}$;

${\displaystyle x^2+x=}$${\displaystyle 12}$:

${\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 3}$;

${\displaystyle x_2=}$${\displaystyle \frac{-1-7}{2}=}$${\displaystyle -4}$;

${\displaystyle x^2+x=}$${\displaystyle -5}$:

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

${\displaystyle a=}$${\displaystyle 1,b=}$${\displaystyle 1,c=}$${\displaystyle 5}$;

${\displaystyle D=}$${\displaystyle b^2-4ac=}$${\displaystyle 1^2-4\cdot 1\cdot 5=}$${\displaystyle 1-20=}$${\displaystyle -19}$;

${\displaystyle D<0}$ — коренів немає;

Корені: ${\displaystyle -4;3}$;

б) ${\displaystyle (2x^2-5x-4)(2x^2-5x)=}$${\displaystyle 21}$;

заміна ${\displaystyle 2x^2-5x=}$${\displaystyle t}$:

${\displaystyle (t-4)t=}$${\displaystyle 21}$;

${\displaystyle t^2-4t-21=}$${\displaystyle 0}$;

${\displaystyle a=}$${\displaystyle 1,b=}$${\displaystyle -4,c=}$${\displaystyle -21}$;

${\displaystyle D=}$${\displaystyle b^2-4ac=}$${\displaystyle {(-4)}^2-4\cdot 1\cdot (-21)=}$${\displaystyle 16+84=}$${\displaystyle 100}$;

${\displaystyle D>0}$, два корені:

${\displaystyle t_{1,2}=}$${\displaystyle \frac{-b\pm \sqrt{D}}{2a}=}$${\displaystyle \frac{4\pm \sqrt{100}}{2}=}$${\displaystyle \frac{4\pm 10}{2}}$;

${\displaystyle t_1=}$${\displaystyle \frac{4+10}{2}=}$${\displaystyle 7}$;

${\displaystyle t_2=}$${\displaystyle \frac{4-10}{2}=}$${\displaystyle -3}$;

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

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

${\displaystyle a=}$${\displaystyle 2,b=}$${\displaystyle -5,c=}$${\displaystyle -7}$;

${\displaystyle D=}$${\displaystyle b^2-4ac=}$${\displaystyle {(-5)}^2-4\cdot 2\cdot (-7)=}$${\displaystyle 25+56=}$${\displaystyle 81}$;

${\displaystyle D>0}$, два корені:

${\displaystyle x_{1,2}=}$${\displaystyle \frac{-b\pm \sqrt{D}}{2a}=}$${\displaystyle \frac{5\pm \sqrt{81}}{4}=}$${\displaystyle \frac{5\pm 9}{4}}$;

${\displaystyle x_1=}$${\displaystyle \frac{5+9}{4}=}$${\displaystyle 3,5}$;

${\displaystyle x_2=}$${\displaystyle \frac{5-9}{4}=}$${\displaystyle -1}$;

${\displaystyle 2x^2-5x=}$${\displaystyle -3}$:

${\displaystyle 2x^2-5x+3=}$${\displaystyle 0}$;

${\displaystyle a=}$${\displaystyle 2,b=}$${\displaystyle -5,c=}$${\displaystyle 3}$;

${\displaystyle D=}$${\displaystyle b^2-4ac=}$${\displaystyle {(-5)}^2-4\cdot 2\cdot 3=}$${\displaystyle 25-24=}$${\displaystyle 1}$;

${\displaystyle D>0}$, два корені:

${\displaystyle x_{1,2}=}$${\displaystyle \frac{-b\pm \sqrt{D}}{2a}=}$${\displaystyle \frac{5\pm \sqrt{1}}{4}=}$${\displaystyle \frac{5\pm 1}{4}}$;

${\displaystyle x_1=}$${\displaystyle \frac{5+1}{4}=}$${\displaystyle 1,5}$;

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

Корені: ${\displaystyle -1;1;1,5;3,5}$;

в) ${\displaystyle x(x+1)(x+2)(x+3)=}$${\displaystyle 24}$;

групуємо: ${\displaystyle \left(x(x+3)\right)\cdot \left((x+1)(x+2)\right)=}$${\displaystyle (x^2+3x)(x^2+3x+2)}$;

${\displaystyle (x^2+3x)(x^2+3x+2)=}$${\displaystyle 24}$;

заміна ${\displaystyle x^2+3x=}$${\displaystyle t}$:

${\displaystyle t(t+2)=}$${\displaystyle 24}$;

${\displaystyle t^2+2t-24=}$${\displaystyle 0}$;

${\displaystyle a=}$${\displaystyle 1,b=}$${\displaystyle 2,c=}$${\displaystyle -24}$;

${\displaystyle D=}$${\displaystyle b^2-4ac=}$${\displaystyle 2^2-4\cdot 1\cdot (-24)=}$${\displaystyle 4+96=}$${\displaystyle 100}$;

${\displaystyle D>0}$, два корені:

${\displaystyle t_{1,2}=}$${\displaystyle \frac{-b\pm \sqrt{D}}{2a}=}$${\displaystyle \frac{-2\pm \sqrt{100}}{2}=}$${\displaystyle \frac{-2\pm 10}{2}}$;

${\displaystyle t_1=}$${\displaystyle \frac{-2+10}{2}=}$${\displaystyle 4}$;

${\displaystyle t_2=}$${\displaystyle \frac{-2-10}{2}=}$${\displaystyle -6}$;

${\displaystyle x^2+3x=}$${\displaystyle 4}$:

${\displaystyle x^2+3x-4=}$${\displaystyle 0}$;

${\displaystyle a=}$${\displaystyle 1,b=}$${\displaystyle 3,c=}$${\displaystyle -4}$;

${\displaystyle D=}$${\displaystyle b^2-4ac=}$${\displaystyle 3^2-4\cdot 1\cdot (-4)=}$${\displaystyle 9+16=}$${\displaystyle 25}$;

${\displaystyle D>0}$, два корені:

${\displaystyle x_{1,2}=}$${\displaystyle \frac{-b\pm \sqrt{D}}{2a}=}$${\displaystyle \frac{-3\pm \sqrt{25}}{2}=}$${\displaystyle \frac{-3\pm 5}{2}}$;

${\displaystyle x_1=}$${\displaystyle \frac{-3+5}{2}=}$${\displaystyle 1}$;

${\displaystyle x_2=}$${\displaystyle \frac{-3-5}{2}=}$${\displaystyle -4}$;

${\displaystyle x^2+3x=}$${\displaystyle -6}$:

${\displaystyle x^2+3x+6=}$${\displaystyle 0}$;

${\displaystyle a=}$${\displaystyle 1,b=}$${\displaystyle 3,c=}$${\displaystyle 6}$;

${\displaystyle D=}$${\displaystyle b^2-4ac=}$${\displaystyle 3^2-4\cdot 1\cdot 6=}$${\displaystyle 9-24=}$${\displaystyle -15}$;

${\displaystyle D<0}$ — коренів немає;

Корені: ${\displaystyle -4;1}$;

г) ${\displaystyle {(x-1)}^2(x^2-2x)=}$${\displaystyle 12}$;

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

${\displaystyle (x^2-2x+1)(x^2-2x)=}$${\displaystyle 12}$;

заміна ${\displaystyle x^2-2x=}$${\displaystyle t}$:

${\displaystyle (t+1)t=}$${\displaystyle 12}$;

${\displaystyle t^2+t-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 t_{1,2}=}$${\displaystyle \frac{-b\pm \sqrt{D}}{2a}=}$${\displaystyle \frac{-1\pm \sqrt{49}}{2}=}$${\displaystyle \frac{-1\pm 7}{2}}$;

${\displaystyle t_1=}$${\displaystyle \frac{-1+7}{2}=}$${\displaystyle 3}$;

${\displaystyle t_2=}$${\displaystyle \frac{-1-7}{2}=}$${\displaystyle -4}$;

${\displaystyle x^2-2x=}$${\displaystyle 3}$:

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

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

${\displaystyle D=}$${\displaystyle b^2-4ac=}$${\displaystyle {(-2)}^2-4\cdot 1\cdot (-3)=}$${\displaystyle 4+12=}$${\displaystyle 16}$;

${\displaystyle D>0}$, два корені:

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

${\displaystyle x_1=}$${\displaystyle \frac{2+4}{2}=}$${\displaystyle 3}$;

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

${\displaystyle x^2-2x=}$${\displaystyle -4}$:

${\displaystyle x^2-2x+4=}$${\displaystyle 0}$;

${\displaystyle a=}$${\displaystyle 1,b=}$${\displaystyle -2,c=}$${\displaystyle 4}$;

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

${\displaystyle D<0}$ — коренів немає;

Корені: ${\displaystyle -1;3}$;

ґ) ${\displaystyle (x^2+2x+1)(x^2+2x+4)=}$${\displaystyle 10}$;

заміна ${\displaystyle x^2+2x=}$${\displaystyle t}$:

${\displaystyle (t+1)(t+4)=}$${\displaystyle 10}$;

${\displaystyle t^2+5t+4-10=}$${\displaystyle 0}$;

${\displaystyle t^2+5t-6=}$${\displaystyle 0}$;

${\displaystyle a=}$${\displaystyle 1,b=}$${\displaystyle 5,c=}$${\displaystyle -6}$;

${\displaystyle D=}$${\displaystyle b^2-4ac=}$${\displaystyle 5^2-4\cdot 1\cdot (-6)=}$${\displaystyle 25+24=}$${\displaystyle 49}$;

${\displaystyle D>0}$, два корені:

${\displaystyle t_{1,2}=}$${\displaystyle \frac{-b\pm \sqrt{D}}{2a}=}$${\displaystyle \frac{-5\pm \sqrt{49}}{2}=}$${\displaystyle \frac{-5\pm 7}{2}}$;

${\displaystyle t_1=}$${\displaystyle \frac{-5+7}{2}=}$${\displaystyle 1}$;

${\displaystyle t_2=}$${\displaystyle \frac{-5-7}{2}=}$${\displaystyle -6}$;

${\displaystyle x^2+2x=}$${\displaystyle 1}$:

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

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

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

${\displaystyle D>0}$, два корені:

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

${\displaystyle x^2+2x=}$${\displaystyle -6}$:

${\displaystyle x^2+2x+6=}$${\displaystyle 0}$;

${\displaystyle a=}$${\displaystyle 1,b=}$${\displaystyle 2,c=}$${\displaystyle 6}$;

${\displaystyle D=}$${\displaystyle b^2-4ac=}$${\displaystyle 2^2-4\cdot 1\cdot 6=}$${\displaystyle 4-24=}$${\displaystyle -20}$;

${\displaystyle D<0}$ — коренів немає;

Корені: ${\displaystyle -1\pm \sqrt{2}}$;

д) ${\displaystyle (x-2)(x+1)(x+2)(x+5)+36=}$${\displaystyle 0}$;

групуємо: ${\displaystyle (x-2)(x+5)=}$${\displaystyle x^2+3x-10}$;

${\displaystyle (x+1)(x+2)=}$${\displaystyle x^2+3x+2}$;

${\displaystyle (x^2+3x-10)(x^2+3x+2)+36=}$${\displaystyle 0}$;

заміна ${\displaystyle x^2+3x=}$${\displaystyle t}$:

${\displaystyle (t-10)(t+2)+36=}$${\displaystyle 0}$;

${\displaystyle t^2-8t-20+36=}$${\displaystyle 0}$;

${\displaystyle t^2-8t+16=}$${\displaystyle 0}$;

${\displaystyle a=}$${\displaystyle 1,b=}$${\displaystyle -8,c=}$${\displaystyle 16}$;

${\displaystyle D=}$${\displaystyle b^2-4ac=}$${\displaystyle {(-8)}^2-4\cdot 1\cdot 16=}$${\displaystyle 64-64=}$${\displaystyle 0}$;

${\displaystyle D=}$${\displaystyle 0}$, один корінь:

${\displaystyle t=}$${\displaystyle -\frac{b}{2a}=}$${\displaystyle \frac{8}{2}=}$${\displaystyle 4}$;

${\displaystyle x^2+3x=}$${\displaystyle 4}$:

${\displaystyle x^2+3x-4=}$${\displaystyle 0}$;

${\displaystyle a=}$${\displaystyle 1,b=}$${\displaystyle 3,c=}$${\displaystyle -4}$;

${\displaystyle D=}$${\displaystyle b^2-4ac=}$${\displaystyle 3^2-4\cdot 1\cdot (-4)=}$${\displaystyle 9+16=}$${\displaystyle 25}$;

${\displaystyle D>0}$, два корені:

${\displaystyle x_{1,2}=}$${\displaystyle \frac{-b\pm \sqrt{D}}{2a}=}$${\displaystyle \frac{-3\pm \sqrt{25}}{2}=}$${\displaystyle \frac{-3\pm 5}{2}}$;

${\displaystyle x_1=}$${\displaystyle \frac{-3+5}{2}=}$${\displaystyle 1}$;

${\displaystyle x_2=}$${\displaystyle \frac{-3-5}{2}=}$${\displaystyle -4}$.