994 Вправа 994Алгебра 8 клас Бевз 2025 · § 19 · стор. 186

994. Знайди корені рівняння. а) |x² + 5x − 3| = 3; б) |2x² − 3x − 1| = 1; в) x² − 10|x| + 24 = 0; г) 6x² + 5|x| − 1 = 0; ґ) x|x| − 7x + 10 = 0; д) 6x² + 7x − 5|x|/x = 0; е) x² − 4√(x²) − 21 = 0; є) x² − 5(√x)² − 6 = 0.

Нагадаємо: ${\displaystyle \left|a\right|=}$${\displaystyle b,b\ge 0}$ ⇔ ${\displaystyle a=}$${\displaystyle b}$ або ${\displaystyle a=}$${\displaystyle -b}$;

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

${\displaystyle x^2=}$${\displaystyle {\left|x\right|}^2}$.

а) ${\displaystyle |x^2+5x-3|=}$${\displaystyle 3}$;

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

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

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

${\displaystyle x_1=}$${\displaystyle \frac{-5+7}{2}=}$${\displaystyle 1}$;

${\displaystyle x_2=}$${\displaystyle \frac{-5-7}{2}=}$${\displaystyle -6}$;

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

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

${\displaystyle x(x+5)=}$${\displaystyle 0}$;

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

${\displaystyle x_4=}$${\displaystyle -5}$;

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

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

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

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

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

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

${\displaystyle D>0}$, два корені:

${\displaystyle x_{1,2}=}$${\displaystyle \frac{-b\pm \sqrt{D}}{2a}=}$${\displaystyle \frac{3\pm \sqrt{25}}{4}=}$${\displaystyle \frac{3\pm 5}{4}}$;

${\displaystyle x_1=}$${\displaystyle \frac{3+5}{4}=}$${\displaystyle 2}$;

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

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

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

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

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

${\displaystyle x_4=}$${\displaystyle 1,5}$;

в) ${\displaystyle x^2-10\left|x\right|+24=}$${\displaystyle 0}$;

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

${\displaystyle t^2-10t+24=}$${\displaystyle 0}$;

${\displaystyle a=}$${\displaystyle 1,b=}$${\displaystyle -10,c=}$${\displaystyle 24}$;

${\displaystyle D=}$${\displaystyle b^2-4ac=}$${\displaystyle {(-10)}^2-4\cdot 1\cdot 24=}$${\displaystyle 100-96=}$${\displaystyle 4}$;

${\displaystyle D>0}$, два корені:

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

${\displaystyle t_1=}$${\displaystyle \frac{10+2}{2}=}$${\displaystyle 6}$;

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

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

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

${\displaystyle \left|x\right|=}$${\displaystyle 4}$;

${\displaystyle x=}$${\displaystyle \pm 4}$;

г) ${\displaystyle 6x^2+5\left|x\right|-1=}$${\displaystyle 0}$;

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

${\displaystyle 6t^2+5t-1=}$${\displaystyle 0}$;

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

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

${\displaystyle D>0}$, два корені:

${\displaystyle t_{1,2}=}$${\displaystyle \frac{-b\pm \sqrt{D}}{2a}=}$${\displaystyle \frac{-5\pm \sqrt{49}}{12}=}$${\displaystyle \frac{-5\pm 7}{12}}$;

${\displaystyle t_1=}$${\displaystyle \frac{-5+7}{12}=}$${\displaystyle \frac{1}{6}}$;

${\displaystyle t_2=}$${\displaystyle \frac{-5-7}{12}=}$${\displaystyle -1}$;

${\displaystyle t=}$${\displaystyle -1<0}$ — не підходить; ${\displaystyle \left|x\right|=}$${\displaystyle \frac{1}{6}}$;

${\displaystyle x=}$${\displaystyle \pm \frac{1}{6}}$;

ґ) ${\displaystyle x\left|x\right|-7x+10=}$${\displaystyle 0}$;

1) ${\displaystyle x\ge 0}$: ${\displaystyle \left|x\right|=}$${\displaystyle x}$:

${\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 5}$;

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

обидва корені ≥${\displaystyle 0}$ — підходять;

2) ${\displaystyle x<0}$: ${\displaystyle \left|x\right|=}$${\displaystyle -x}$:

${\displaystyle -x^2-7x+10=}$${\displaystyle 0\mid \cdot \left(-1\right)}$;

${\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 89}$;

${\displaystyle D>0}$, два корені:

${\displaystyle x_{1,2}=}$${\displaystyle \frac{-b\pm \sqrt{D}}{2a}=}$${\displaystyle \frac{-7\pm \sqrt{89}}{2}}$;

${\displaystyle x<0}$, тому підходить лише ${\displaystyle x=}$${\displaystyle \frac{-7-\sqrt{89}}{2}}$;

д) ${\displaystyle 6x^2+7x-5\cdot \frac{\left|x\right|}{x}=}$${\displaystyle 0}$, ${\displaystyle x\ne 0}$;

1) ${\displaystyle x>0}$: ${\displaystyle \frac{\left|x\right|}{x}=}$${\displaystyle 1}$:

${\displaystyle 6x^2+7x-5=}$${\displaystyle 0}$;

${\displaystyle a=}$${\displaystyle 6,b=}$${\displaystyle 7,c=}$${\displaystyle -5}$;

${\displaystyle D=}$${\displaystyle b^2-4ac=}$${\displaystyle 7^2-4\cdot 6\cdot (-5)=}$${\displaystyle 49+120=}$${\displaystyle 169}$;

${\displaystyle D>0}$, два корені:

${\displaystyle x_{1,2}=}$${\displaystyle \frac{-b\pm \sqrt{D}}{2a}=}$${\displaystyle \frac{-7\pm \sqrt{169}}{12}=}$${\displaystyle \frac{-7\pm 13}{12}}$;

${\displaystyle x_1=}$${\displaystyle \frac{-7+13}{12}=}$${\displaystyle 0,5}$;

${\displaystyle x_2=}$${\displaystyle \frac{-7-13}{12}=}$${\displaystyle -\frac{5}{3}}$;

${\displaystyle x>0}$, тому ${\displaystyle x=}$${\displaystyle 0,5}$;

2) ${\displaystyle x<0}$: ${\displaystyle \frac{\left|x\right|}{x}=}$${\displaystyle -1}$:

${\displaystyle 6x^2+7x+5=}$${\displaystyle 0}$;

${\displaystyle a=}$${\displaystyle 6,b=}$${\displaystyle 7,c=}$${\displaystyle 5}$;

${\displaystyle D=}$${\displaystyle b^2-4ac=}$${\displaystyle 7^2-4\cdot 6\cdot 5=}$${\displaystyle 49-120=}$${\displaystyle -71}$;

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

е) ${\displaystyle x^2-4\sqrt{x^2}-21=}$${\displaystyle 0}$;

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

${\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 t=}$${\displaystyle -3<0}$ — не підходить; ${\displaystyle \left|x\right|=}$${\displaystyle 7}$;

${\displaystyle x=}$${\displaystyle \pm 7}$;

є) ${\displaystyle x^2-5{\left(\sqrt{x}\right)}^2-6=}$${\displaystyle 0}$;

вираз ${\displaystyle \sqrt{x}}$ має зміст при ${\displaystyle x\ge 0}$, тоді ${\displaystyle {\left(\sqrt{x}\right)}^2=}$${\displaystyle x}$:

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

${\displaystyle x_1=}$${\displaystyle \frac{5+7}{2}=}$${\displaystyle 6}$;

${\displaystyle x_2=}$${\displaystyle \frac{5-7}{2}=}$${\displaystyle -1}$;

${\displaystyle x=}$${\displaystyle -1<0}$ — не підходить; ${\displaystyle x=}$${\displaystyle 6}$.