971 Вправа 971Алгебра 8 клас Бевз 2025 · § 19 · стор. 184

971. Розв’яжи рівняння. а) 15/(x + 2) = x; б) 3c²/(c − 1) = c − 2; в) (x − 5)/(x + 3) = (3 + 2x)/(1 − 6x); г) (2x − 1)/(3 − 2x) = (4x + 1)/(2x + 3); ґ) (x² + x − 12)/(x + 4) = 0; д) (4x² − 11x − 3)/(x² − 9) = 0.

а) ${\displaystyle \frac{15}{x+2}=}$${\displaystyle x\mid \cdot \left(x+2\right)}$;

ОДЗ: ${\displaystyle x\ne -2}$;

${\displaystyle 15=}$${\displaystyle x(x+2)}$;

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

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

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

${\displaystyle a=}$${\displaystyle 1,b=}$${\displaystyle 2,c=}$${\displaystyle -15}$;

${\displaystyle D=}$${\displaystyle b^2-4ac=}$${\displaystyle 2^2-4\cdot 1\cdot (-15)=}$${\displaystyle 4+60=}$${\displaystyle 64}$;

${\displaystyle D>0}$, два корені:

${\displaystyle x_{1,2}=}$${\displaystyle \frac{-b\pm \sqrt{D}}{2a}=}$${\displaystyle \frac{-2\pm \sqrt{64}}{2}=}$${\displaystyle \frac{-2\pm 8}{2}}$;

${\displaystyle x_1=}$${\displaystyle \frac{-2+8}{2}=}$${\displaystyle 3}$;

${\displaystyle x_2=}$${\displaystyle \frac{-2-8}{2}=}$${\displaystyle -5}$;

Отже, ${\displaystyle x=}$${\displaystyle 3\text{або }x=}$${\displaystyle -5}$;

б) ${\displaystyle \frac{3c^2}{c-1}=}$${\displaystyle c-2\mid \cdot \left(c-1\right)}$;

ОДЗ: ${\displaystyle c\ne 1}$;

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

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

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

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

${\displaystyle c_1=}$${\displaystyle \frac{-3+5}{4}=}$${\displaystyle 0,5}$;

${\displaystyle c_2=}$${\displaystyle \frac{-3-5}{4}=}$${\displaystyle -2}$;

Отже, ${\displaystyle c=}$${\displaystyle 0,5\text{або }c=}$${\displaystyle -2}$;

в) ${\displaystyle \frac{x-5}{x+3}=}$${\displaystyle \frac{3+2x}{1-6x}\mid \cdot \left(x+3\right)\left(1-6x\right)}$;

ОДЗ: ${\displaystyle x\ne -3,x\ne \frac{1}{6}}$;

${\displaystyle (x-5)(1-6x)=}$${\displaystyle (3+2x)(x+3)}$;

${\displaystyle x-6x^2-5+30x=}$${\displaystyle 3x+9+2x^2+6x}$;

${\displaystyle -6x^2+31x-5=}$${\displaystyle 2x^2+9x+9}$;

${\displaystyle -8x^2+22x-14=}$${\displaystyle 0\mid :\left(-2\right)}$;

${\displaystyle 4x^2-11x+7=}$${\displaystyle 0}$;

${\displaystyle a=}$${\displaystyle 4,b=}$${\displaystyle -11,c=}$${\displaystyle 7}$;

${\displaystyle D=}$${\displaystyle b^2-4ac=}$${\displaystyle {(-11)}^2-4\cdot 4\cdot 7=}$${\displaystyle 121-112=}$${\displaystyle 9}$;

${\displaystyle D>0}$, два корені:

${\displaystyle x_{1,2}=}$${\displaystyle \frac{-b\pm \sqrt{D}}{2a}=}$${\displaystyle \frac{11\pm \sqrt{9}}{8}=}$${\displaystyle \frac{11\pm 3}{8}}$;

${\displaystyle x_1=}$${\displaystyle \frac{11+3}{8}=}$${\displaystyle 1,75}$;

${\displaystyle x_2=}$${\displaystyle \frac{11-3}{8}=}$${\displaystyle 1}$;

Отже, ${\displaystyle x=}$${\displaystyle 1,75\text{або }x=}$${\displaystyle 1}$;

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

ОДЗ: ${\displaystyle x\ne 1,5,x\ne -1,5}$;

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

${\displaystyle 4x^2+6x-2x-3=}$${\displaystyle 12x-8x^2+3-2x}$;

${\displaystyle 4x^2+4x-3=}$${\displaystyle -8x^2+10x+3}$;

${\displaystyle 12x^2-6x-6=}$${\displaystyle 0\mid :6}$;

${\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}$;

Отже, ${\displaystyle x=}$${\displaystyle 1\text{або }x=}$${\displaystyle -0,5}$;

ґ) ${\displaystyle \frac{x^2+x-12}{x+4}=}$${\displaystyle 0}$;

ОДЗ: ${\displaystyle x\ne -4}$;

дріб дорівнює нулю, якщо чисельник дорівнює нулю, а знаменник — ні:

${\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}$ — не входить в ОДЗ (сторонній корінь);

Отже, ${\displaystyle x=}$${\displaystyle 3}$;

д) ${\displaystyle \frac{4x^2-11x-3}{x^2-9}=}$${\displaystyle 0}$;

ОДЗ: ${\displaystyle x\ne \pm 3}$;

дріб дорівнює нулю, якщо чисельник дорівнює нулю, а знаменник — ні:

${\displaystyle 4x^2-11x-3=}$${\displaystyle 0}$;

${\displaystyle a=}$${\displaystyle 4,b=}$${\displaystyle -11,c=}$${\displaystyle -3}$;

${\displaystyle D=}$${\displaystyle b^2-4ac=}$${\displaystyle {(-11)}^2-4\cdot 4\cdot (-3)=}$${\displaystyle 121+48=}$${\displaystyle 169}$;

${\displaystyle D>0}$, два корені:

${\displaystyle x_{1,2}=}$${\displaystyle \frac{-b\pm \sqrt{D}}{2a}=}$${\displaystyle \frac{11\pm \sqrt{169}}{8}=}$${\displaystyle \frac{11\pm 13}{8}}$;

${\displaystyle x_1=}$${\displaystyle \frac{11+13}{8}=}$${\displaystyle 3}$;

${\displaystyle x_2=}$${\displaystyle \frac{11-13}{8}=}$${\displaystyle -0,25}$;

${\displaystyle x=}$${\displaystyle 3}$ — не входить в ОДЗ (сторонній корінь);

Отже, ${\displaystyle x=}$${\displaystyle -0,25}$.