989 Вправа 989Алгебра 8 клас Бевз 2025 · § 19 · стор. 185

989. Розв’яжи рівняння. а) (x² − 2x)/(x + 2) = 8/(x + 2); б) (x² − 2x)/(x − 3) = (5x − 12)/(x − 3); в) (c − 7)/(c + 5) + (c + 4)/(5 − c) = 3; г) (2x + 3)/(2x − 3) − (3x − 1)/(x − 1) = 2; ґ) 7/(x + 2) − 3/(2 − x) = 16/x; д) (4x − 2)/(x + 4) − (3x − 14)/(x − 4) = (2x − 1)/(x² − 16).

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

ОДЗ: ${\displaystyle x\ne -2}$;

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

${\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}$;

${\displaystyle x=}$${\displaystyle -2}$ — не входить в ОДЗ (сторонній корінь);

Отже, ${\displaystyle x=}$${\displaystyle 4}$;

б) ${\displaystyle \frac{x^2-2x}{x-3}=}$${\displaystyle \frac{5x-12}{x-3}\mid \cdot \left(x-3\right)}$;

ОДЗ: ${\displaystyle x\ne 3}$;

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

${\displaystyle x^2-2x-5x+12=}$${\displaystyle 0}$;

${\displaystyle x^2-7x+12=}$${\displaystyle 0}$;

${\displaystyle a=}$${\displaystyle 1,b=}$${\displaystyle -7,c=}$${\displaystyle 12}$;

${\displaystyle D=}$${\displaystyle b^2-4ac=}$${\displaystyle {(-7)}^2-4\cdot 1\cdot 12=}$${\displaystyle 49-48=}$${\displaystyle 1}$;

${\displaystyle D>0}$, два корені:

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

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

${\displaystyle x_2=}$${\displaystyle \frac{7-1}{2}=}$${\displaystyle 3}$;

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

Отже, ${\displaystyle x=}$${\displaystyle 4}$;

в) ${\displaystyle \frac{c-7}{c+5}+\frac{c+4}{5-c}=}$${\displaystyle 3\mid \cdot \left(c+5\right)\left(5-c\right)}$;

ОДЗ: ${\displaystyle c\ne \pm 5}$;

${\displaystyle (c-7)(5-c)+(c+4)(c+5)=}$${\displaystyle 3(c+5)(5-c)}$;

${\displaystyle -c^2+12c-35+c^2+9c+20=}$${\displaystyle 3(25-c^2)}$;

${\displaystyle 21c-15=}$${\displaystyle 75-3c^2}$;

${\displaystyle 3c^2+21c-90=}$${\displaystyle 0\mid :3}$;

${\displaystyle c^2+7c-30=}$${\displaystyle 0}$;

${\displaystyle a=}$${\displaystyle 1,b=}$${\displaystyle 7,c=}$${\displaystyle -30}$;

${\displaystyle D=}$${\displaystyle b^2-4ac=}$${\displaystyle 7^2-4\cdot 1\cdot (-30)=}$${\displaystyle 49+120=}$${\displaystyle 169}$;

${\displaystyle D>0}$, два корені:

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

${\displaystyle c_1=}$${\displaystyle \frac{-7+13}{2}=}$${\displaystyle 3}$;

${\displaystyle c_2=}$${\displaystyle \frac{-7-13}{2}=}$${\displaystyle -10}$;

Отже, ${\displaystyle c=}$${\displaystyle 3\text{або }c=}$${\displaystyle -10}$;

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

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

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

${\displaystyle 2x^2+x-3-(6x^2-11x+3)=}$${\displaystyle 2(2x^2-5x+3)}$;

${\displaystyle -4x^2+12x-6=}$${\displaystyle 4x^2-10x+6}$;

${\displaystyle -8x^2+22x-12=}$${\displaystyle 0\mid :\left(-2\right)}$;

${\displaystyle 4x^2-11x+6=}$${\displaystyle 0}$;

${\displaystyle a=}$${\displaystyle 4,b=}$${\displaystyle -11,c=}$${\displaystyle 6}$;

${\displaystyle D=}$${\displaystyle b^2-4ac=}$${\displaystyle {(-11)}^2-4\cdot 4\cdot 6=}$${\displaystyle 121-96=}$${\displaystyle 25}$;

${\displaystyle D>0}$, два корені:

${\displaystyle x_{1,2}=}$${\displaystyle \frac{-b\pm \sqrt{D}}{2a}=}$${\displaystyle \frac{11\pm \sqrt{25}}{8}=}$${\displaystyle \frac{11\pm 5}{8}}$;

${\displaystyle x_1=}$${\displaystyle \frac{11+5}{8}=}$${\displaystyle 2}$;

${\displaystyle x_2=}$${\displaystyle \frac{11-5}{8}=}$${\displaystyle 0,75}$;

Отже, ${\displaystyle x=}$${\displaystyle 2\text{або }x=}$${\displaystyle 0,75}$;

ґ) ${\displaystyle \frac{7}{x+2}-\frac{3}{2-x}=}$${\displaystyle \frac{16}{x}\mid \cdot \left(x\left(x+2\right)\left(2-x\right)\right)}$;

ОДЗ: ${\displaystyle x\ne 0,x\ne \pm 2}$;

${\displaystyle 7x(2-x)-3x(x+2)=}$${\displaystyle 16(x+2)(2-x)}$;

${\displaystyle 14x-7x^2-3x^2-6x=}$${\displaystyle 16(4-x^2)}$;

${\displaystyle -10x^2+8x=}$${\displaystyle 64-16x^2}$;

${\displaystyle 6x^2+8x-64=}$${\displaystyle 0\mid :2}$;

${\displaystyle 3x^2+4x-32=}$${\displaystyle 0}$;

${\displaystyle a=}$${\displaystyle 3,b=}$${\displaystyle 4,c=}$${\displaystyle -32}$;

${\displaystyle D=}$${\displaystyle b^2-4ac=}$${\displaystyle 4^2-4\cdot 3\cdot (-32)=}$${\displaystyle 16+384=}$${\displaystyle 400}$;

${\displaystyle D>0}$, два корені:

${\displaystyle x_{1,2}=}$${\displaystyle \frac{-b\pm \sqrt{D}}{2a}=}$${\displaystyle \frac{-4\pm \sqrt{400}}{6}=}$${\displaystyle \frac{-4\pm 20}{6}}$;

${\displaystyle x_1=}$${\displaystyle \frac{-4+20}{6}=}$${\displaystyle \frac{8}{3}}$;

${\displaystyle x_2=}$${\displaystyle \frac{-4-20}{6}=}$${\displaystyle -4}$;

Отже, ${\displaystyle x=}$${\displaystyle \frac{8}{3}\text{або }x=}$${\displaystyle -4}$;

д) ${\displaystyle \frac{4x-2}{x+4}-\frac{3x-14}{x-4}=}$${\displaystyle \frac{2x-1}{x^2-16}\mid \cdot \left(x+4\right)\left(x-4\right)}$;

ОДЗ: ${\displaystyle x\ne \pm 4}$;

${\displaystyle (4x-2)(x-4)-(3x-14)(x+4)=}$${\displaystyle 2x-1}$;

${\displaystyle 4x^2-18x+8-(3x^2-2x-56)=}$${\displaystyle 2x-1}$;

${\displaystyle x^2-16x+64=}$${\displaystyle 2x-1}$;

${\displaystyle x^2-16x+64-2x+1=}$${\displaystyle 0}$;

${\displaystyle x^2-18x+65=}$${\displaystyle 0}$;

${\displaystyle a=}$${\displaystyle 1,b=}$${\displaystyle -18,c=}$${\displaystyle 65}$;

${\displaystyle D=}$${\displaystyle b^2-4ac=}$${\displaystyle {(-18)}^2-4\cdot 1\cdot 65=}$${\displaystyle 324-260=}$${\displaystyle 64}$;

${\displaystyle D>0}$, два корені:

${\displaystyle x_{1,2}=}$${\displaystyle \frac{-b\pm \sqrt{D}}{2a}=}$${\displaystyle \frac{18\pm \sqrt{64}}{2}=}$${\displaystyle \frac{18\pm 8}{2}}$;

${\displaystyle x_1=}$${\displaystyle \frac{18+8}{2}=}$${\displaystyle 13}$;

${\displaystyle x_2=}$${\displaystyle \frac{18-8}{2}=}$${\displaystyle 5}$;

Отже, ${\displaystyle x=}$${\displaystyle 13\text{або }x=}$${\displaystyle 5}$.