991 Вправа 991Алгебра 8 клас Бевз 2025 · § 19 · стор. 186

991. Розв’яжи рівняння. а) 2(x + 7)/(x − 1) − (3x + 7)/(x² − 1) + (x − 1)/(x + 1) = 4; б) (x + 1)/(4x) − (4x − 1)/(3x² − 6x) = (x − 4)/(2x − 4); в) (9 − 2x)/(x² − 3x) + (4 − x)/(x² − 6x + 9) + 1/x = 0; г) 4/(x + 1) − (2x − 1)/(x² − x + 1) − (3x² − 1)/(x³ + 1) = 0.

а) ${\displaystyle \frac{2(x+7)}{x-1}-\frac{3x+7}{(x-1)(x+1)}+\frac{x-1}{x+1}=}$${\displaystyle 4\mid \cdot \left(x-1\right)\left(x+1\right)}$;

ОДЗ: ${\displaystyle x\ne \pm 1}$;

${\displaystyle 2(x+7)(x+1)-(3x+7)+{(x-1)}^2=}$${\displaystyle 4(x^2-1)}$;

${\displaystyle 2x^2+16x+14-3x-7+x^2-2x+1=}$${\displaystyle 4x^2-4}$;

${\displaystyle 3x^2+11x+8=}$${\displaystyle 4x^2-4}$;

${\displaystyle -x^2+11x+12=}$${\displaystyle 0}$;

${\displaystyle x^2-11x-12=}$${\displaystyle 0}$;

${\displaystyle a=}$${\displaystyle 1,b=}$${\displaystyle -11,c=}$${\displaystyle -12}$;

${\displaystyle D=}$${\displaystyle b^2-4ac=}$${\displaystyle {(-11)}^2-4\cdot 1\cdot (-12)=}$${\displaystyle 121+48=}$${\displaystyle 169}$;

${\displaystyle D>0}$, два корені:

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

${\displaystyle x_1=}$${\displaystyle \frac{11+13}{2}=}$${\displaystyle 12}$;

${\displaystyle x_2=}$${\displaystyle \frac{11-13}{2}=}$${\displaystyle -1}$;

${\displaystyle x=}$${\displaystyle -1}$ — не входить в ОДЗ (сторонній корінь);

Отже, ${\displaystyle x=}$${\displaystyle 12}$;

б) ${\displaystyle \frac{x+1}{4x}-\frac{4x-1}{3x(x-2)}=}$${\displaystyle \frac{x-4}{2(x-2)}\mid \cdot \left(12x\left(x-2\right)\right)}$;

ОДЗ: ${\displaystyle x\ne 0,x\ne 2}$;

${\displaystyle 3(x+1)(x-2)-4(4x-1)=}$${\displaystyle 6x(x-4)}$;

${\displaystyle 3x^2-3x-6-16x+4=}$${\displaystyle 6x^2-24x}$;

${\displaystyle 3x^2-19x-2-6x^2+24x=}$${\displaystyle 0}$;

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

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

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

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

${\displaystyle D>0}$, два корені:

${\displaystyle x_{1,2}=}$${\displaystyle \frac{-b\pm \sqrt{D}}{2a}=}$${\displaystyle \frac{5\pm \sqrt{1}}{6}=}$${\displaystyle \frac{5\pm 1}{6}}$;

${\displaystyle x_1=}$${\displaystyle \frac{5+1}{6}=}$${\displaystyle 1}$;

${\displaystyle x_2=}$${\displaystyle \frac{5-1}{6}=}$${\displaystyle \frac{2}{3}}$;

Отже, ${\displaystyle x=}$${\displaystyle 1\text{або }x=}$${\displaystyle \frac{2}{3}}$;

в) ${\displaystyle \frac{9-2x}{x(x-3)}+\frac{4-x}{{(x-3)}^2}+\frac{1}{x}=}$${\displaystyle 0\mid \cdot \left(x{\left(x-3\right)}^2\right)}$;

ОДЗ: ${\displaystyle x\ne 0,x\ne 3}$;

${\displaystyle (9-2x)(x-3)+x(4-x)+{(x-3)}^2=}$${\displaystyle 0}$;

${\displaystyle -2x^2+15x-27+4x-x^2+x^2-6x+9=}$${\displaystyle 0}$;

${\displaystyle -2x^2+13x-18=}$${\displaystyle 0}$;

${\displaystyle 2x^2-13x+18=}$${\displaystyle 0}$;

${\displaystyle a=}$${\displaystyle 2,b=}$${\displaystyle -13,c=}$${\displaystyle 18}$;

${\displaystyle D=}$${\displaystyle b^2-4ac=}$${\displaystyle {(-13)}^2-4\cdot 2\cdot 18=}$${\displaystyle 169-144=}$${\displaystyle 25}$;

${\displaystyle D>0}$, два корені:

${\displaystyle x_{1,2}=}$${\displaystyle \frac{-b\pm \sqrt{D}}{2a}=}$${\displaystyle \frac{13\pm \sqrt{25}}{4}=}$${\displaystyle \frac{13\pm 5}{4}}$;

${\displaystyle x_1=}$${\displaystyle \frac{13+5}{4}=}$${\displaystyle 4,5}$;

${\displaystyle x_2=}$${\displaystyle \frac{13-5}{4}=}$${\displaystyle 2}$;

Отже, ${\displaystyle x=}$${\displaystyle 4,5\text{або }x=}$${\displaystyle 2}$;

(тут ${\displaystyle x^3+1=}$${\displaystyle (x+1)(x^2-x+1)}$, ${\displaystyle x^2-x+1>0}$)

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

ОДЗ: ${\displaystyle x\ne -1}$;

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

${\displaystyle 4x^2-4x+4-2x^2-x+1-3x^2+1=}$${\displaystyle 0}$;

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

${\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}$;

Отже, ${\displaystyle x=}$${\displaystyle 1\text{або }x=}$${\displaystyle -6}$.