995 Вправа 995Алгебра 8 клас Бевз 2025 · § 19 · стор. 186

995. Знайди корені рівняння. а) |x² + 7x + 5| = 5; б) |2x² − x − 3| = 3; в) x² − 8|x| + 12 = 0; г) 3x² − 4|x| − 20 = 0; ґ) x|x| − 4x + 3 = 0; д) 2x² − 7x − 4|x|/x = 0; е) x² + 2√(x²) − 3 = 0; є) x² + (√x)² − 12 = 0.

а) ${\displaystyle |x^2+7x+5|=}$${\displaystyle 5}$;

1) ${\displaystyle x^2+7x+5=}$${\displaystyle 5}$;

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

${\displaystyle x(x+7)=}$${\displaystyle 0}$;

${\displaystyle x_1=}$${\displaystyle 0}$;

${\displaystyle x_2=}$${\displaystyle -7}$;

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

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

${\displaystyle x^2+7x+10=}$${\displaystyle 0}$;

${\displaystyle a=}$${\displaystyle 1,b=}$${\displaystyle 7,c=}$${\displaystyle 10}$;

${\displaystyle D=}$${\displaystyle b^2-4ac=}$${\displaystyle 7^2-4\cdot 1\cdot 10=}$${\displaystyle 49-40=}$${\displaystyle 9}$;

${\displaystyle D>0}$, два корені:

${\displaystyle x_{1,2}=}$${\displaystyle \frac{-b\pm \sqrt{D}}{2a}=}$${\displaystyle \frac{-7\pm \sqrt{9}}{2}=}$${\displaystyle \frac{-7\pm 3}{2}}$;

${\displaystyle x_1=}$${\displaystyle \frac{-7+3}{2}=}$${\displaystyle -2}$;

${\displaystyle x_2=}$${\displaystyle \frac{-7-3}{2}=}$${\displaystyle -5}$;

б) ${\displaystyle |2x^2-x-3|=}$${\displaystyle 3}$;

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

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

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

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

${\displaystyle D=}$${\displaystyle b^2-4ac=}$${\displaystyle {(-1)}^2-4\cdot 2\cdot (-6)=}$${\displaystyle 1+48=}$${\displaystyle 49}$;

${\displaystyle D>0}$, два корені:

${\displaystyle x_{1,2}=}$${\displaystyle \frac{-b\pm \sqrt{D}}{2a}=}$${\displaystyle \frac{1\pm \sqrt{49}}{4}=}$${\displaystyle \frac{1\pm 7}{4}}$;

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

${\displaystyle x_2=}$${\displaystyle \frac{1-7}{4}=}$${\displaystyle -1,5}$;

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

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

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

${\displaystyle x_3=}$${\displaystyle 0}$;

${\displaystyle x_4=}$${\displaystyle 0,5}$;

в) ${\displaystyle x^2-8\left|x\right|+12=}$${\displaystyle 0}$;

заміна ${\displaystyle \left|x\right|=}$${\displaystyle t}$, ${\displaystyle t\ge 0}$:

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

${\displaystyle t_1=}$${\displaystyle \frac{8+4}{2}=}$${\displaystyle 6}$;

${\displaystyle t_2=}$${\displaystyle \frac{8-4}{2}=}$${\displaystyle 2}$;

${\displaystyle \left|x\right|=}$${\displaystyle 6}$;

${\displaystyle x=}$${\displaystyle \pm 6}$;

${\displaystyle \left|x\right|=}$${\displaystyle 2}$;

${\displaystyle x=}$${\displaystyle \pm 2}$;

г) ${\displaystyle 3x^2-4\left|x\right|-20=}$${\displaystyle 0}$;

заміна ${\displaystyle \left|x\right|=}$${\displaystyle t}$, ${\displaystyle t\ge 0}$:

${\displaystyle 3t^2-4t-20=}$${\displaystyle 0}$;

${\displaystyle a=}$${\displaystyle 3,b=}$${\displaystyle -4,c=}$${\displaystyle -20}$;

${\displaystyle D=}$${\displaystyle b^2-4ac=}$${\displaystyle {(-4)}^2-4\cdot 3\cdot (-20)=}$${\displaystyle 16+240=}$${\displaystyle 256}$;

${\displaystyle D>0}$, два корені:

${\displaystyle t_{1,2}=}$${\displaystyle \frac{-b\pm \sqrt{D}}{2a}=}$${\displaystyle \frac{4\pm \sqrt{256}}{6}=}$${\displaystyle \frac{4\pm 16}{6}}$;

${\displaystyle t_1=}$${\displaystyle \frac{4+16}{6}=}$${\displaystyle \frac{10}{3}}$;

${\displaystyle t_2=}$${\displaystyle \frac{4-16}{6}=}$${\displaystyle -2}$;

${\displaystyle t=}$${\displaystyle -2<0}$ — не підходить; ${\displaystyle \left|x\right|=}$${\displaystyle \frac{10}{3}}$;

${\displaystyle x=}$${\displaystyle \pm 3\frac{1}{3}}$;

ґ) ${\displaystyle x\left|x\right|-4x+3=}$${\displaystyle 0}$;

1) ${\displaystyle x\ge 0}$:

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

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

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

${\displaystyle D>0}$, два корені:

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

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

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

обидва корені підходять;

2) ${\displaystyle x<0}$:

${\displaystyle -x^2-4x+3=}$${\displaystyle 0\mid \cdot \left(-1\right)}$;

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

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

${\displaystyle D=}$${\displaystyle b^2-4ac=}$${\displaystyle 4^2-4\cdot 1\cdot (-3)=}$${\displaystyle 16+12=}$${\displaystyle 28}$;

${\displaystyle D>0}$, два корені:

${\displaystyle x_{1,2}=}$${\displaystyle \frac{-b\pm \sqrt{D}}{2a}=}$${\displaystyle \frac{-4\pm \sqrt{28}}{2}=}$${\displaystyle \frac{-4\pm 2\sqrt{7}}{2}=}$${\displaystyle -2\pm \sqrt{7}}$;

${\displaystyle x<0}$, тому підходить лише ${\displaystyle x=}$${\displaystyle -2-\sqrt{7}}$;

д) ${\displaystyle 2x^2-7x-4\cdot \frac{\left|x\right|}{x}=}$${\displaystyle 0}$, ${\displaystyle x\ne 0}$;

1) ${\displaystyle x>0}$:

${\displaystyle 2x^2-7x-4=}$${\displaystyle 0}$;

${\displaystyle a=}$${\displaystyle 2,b=}$${\displaystyle -7,c=}$${\displaystyle -4}$;

${\displaystyle D=}$${\displaystyle b^2-4ac=}$${\displaystyle {(-7)}^2-4\cdot 2\cdot (-4)=}$${\displaystyle 49+32=}$${\displaystyle 81}$;

${\displaystyle D>0}$, два корені:

${\displaystyle x_{1,2}=}$${\displaystyle \frac{-b\pm \sqrt{D}}{2a}=}$${\displaystyle \frac{7\pm \sqrt{81}}{4}=}$${\displaystyle \frac{7\pm 9}{4}}$;

${\displaystyle x_1=}$${\displaystyle \frac{7+9}{4}=}$${\displaystyle 4}$;

${\displaystyle x_2=}$${\displaystyle \frac{7-9}{4}=}$${\displaystyle -0,5}$;

${\displaystyle x>0}$, тому ${\displaystyle x=}$${\displaystyle 4}$;

2) ${\displaystyle x<0}$:

${\displaystyle 2x^2-7x+4=}$${\displaystyle 0}$;

${\displaystyle a=}$${\displaystyle 2,b=}$${\displaystyle -7,c=}$${\displaystyle 4}$;

${\displaystyle D=}$${\displaystyle b^2-4ac=}$${\displaystyle {(-7)}^2-4\cdot 2\cdot 4=}$${\displaystyle 49-32=}$${\displaystyle 17}$;

${\displaystyle D>0}$, два корені:

${\displaystyle x_{1,2}=}$${\displaystyle \frac{-b\pm \sqrt{D}}{2a}=}$${\displaystyle \frac{7\pm \sqrt{17}}{4}}$;

обидва корені додатні — при ${\displaystyle x<0}$ розв’язків немає;

е) ${\displaystyle x^2+2\sqrt{x^2}-3=}$${\displaystyle 0}$;

${\displaystyle \sqrt{x^2}=}$${\displaystyle \left|x\right|=}$${\displaystyle t}$, ${\displaystyle t\ge 0}$:

${\displaystyle t^2+2t-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 t_{1,2}=}$${\displaystyle \frac{-b\pm \sqrt{D}}{2a}=}$${\displaystyle \frac{-2\pm \sqrt{16}}{2}=}$${\displaystyle \frac{-2\pm 4}{2}}$;

${\displaystyle t_1=}$${\displaystyle \frac{-2+4}{2}=}$${\displaystyle 1}$;

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

${\displaystyle t=}$${\displaystyle -3<0}$ — не підходить; ${\displaystyle \left|x\right|=}$${\displaystyle 1}$;

${\displaystyle x=}$${\displaystyle \pm 1}$;

є) ${\displaystyle x^2+{\left(\sqrt{x}\right)}^2-12=}$${\displaystyle 0}$, ${\displaystyle x\ge 0}$;

${\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 3}$;

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

${\displaystyle x=}$${\displaystyle -4<0}$ — не підходить; ${\displaystyle x=}$${\displaystyle 3}$.