990 Вправа 990Алгебра 8 клас Бевз 2025 · § 19 · стор. 185

990. Розв’яжи рівняння. а) 3x/(2x − 1) + (x − 1)/(2x + 1) − (4x − 2)/(4x² − 1) = 5; б) (x + 3)/(2x) − (2x − 1)/(x² − 3x) = (3x − 17)/(4x − 12); в) (2x − 1)/(x² − 4x + 4) − (5x − 6)/(x² − 2x) = 6/x; г) 1/(x − 1) − (8x − 2)/(x² + x + 1) = (x − 9)/(x³ − 1).

Розкладаємо знаменники на множники і знаходимо спільний знаменник.

а) ${\displaystyle \frac{3x}{2x-1}+\frac{x-1}{2x+1}-\frac{4x-2}{(2x-1)(2x+1)}=}$${\displaystyle 5\mid \cdot \left(2x-1\right)\left(2x+1\right)}$;

ОДЗ: ${\displaystyle x\ne \pm 0,5}$;

${\displaystyle 3x(2x+1)+(x-1)(2x-1)-(4x-2)=}$${\displaystyle 5(4x^2-1)}$;

${\displaystyle 6x^2+3x+2x^2-3x+1-4x+2=}$${\displaystyle 20x^2-5}$;

${\displaystyle 8x^2-4x+3=}$${\displaystyle 20x^2-5}$;

${\displaystyle -12x^2-4x+8=}$${\displaystyle 0\mid :\left(-4\right)}$;

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

${\displaystyle a=}$${\displaystyle 3,b=}$${\displaystyle 1,c=}$${\displaystyle -2}$;

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

${\displaystyle D>0}$, два корені:

${\displaystyle x_{1,2}=}$${\displaystyle \frac{-b\pm \sqrt{D}}{2a}=}$${\displaystyle \frac{-1\pm \sqrt{25}}{6}=}$${\displaystyle \frac{-1\pm 5}{6}}$;

${\displaystyle x_1=}$${\displaystyle \frac{-1+5}{6}=}$${\displaystyle \frac{2}{3}}$;

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

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

б) ${\displaystyle \frac{x+3}{2x}-\frac{2x-1}{x(x-3)}=}$${\displaystyle \frac{3x-17}{4(x-3)}\mid \cdot \left(4x\left(x-3\right)\right)}$;

ОДЗ: ${\displaystyle x\ne 0,x\ne 3}$;

${\displaystyle 2(x+3)(x-3)-4(2x-1)=}$${\displaystyle x(3x-17)}$;

${\displaystyle 2x^2-18-8x+4=}$${\displaystyle 3x^2-17x}$;

${\displaystyle 2x^2-8x-14-3x^2+17x=}$${\displaystyle 0}$;

${\displaystyle -x^2+9x-14=}$${\displaystyle 0}$;

${\displaystyle x^2-9x+14=}$${\displaystyle 0}$;

${\displaystyle a=}$${\displaystyle 1,b=}$${\displaystyle -9,c=}$${\displaystyle 14}$;

${\displaystyle D=}$${\displaystyle b^2-4ac=}$${\displaystyle {(-9)}^2-4\cdot 1\cdot 14=}$${\displaystyle 81-56=}$${\displaystyle 25}$;

${\displaystyle D>0}$, два корені:

${\displaystyle x_{1,2}=}$${\displaystyle \frac{-b\pm \sqrt{D}}{2a}=}$${\displaystyle \frac{9\pm \sqrt{25}}{2}=}$${\displaystyle \frac{9\pm 5}{2}}$;

${\displaystyle x_1=}$${\displaystyle \frac{9+5}{2}=}$${\displaystyle 7}$;

${\displaystyle x_2=}$${\displaystyle \frac{9-5}{2}=}$${\displaystyle 2}$;

Отже, ${\displaystyle x=}$${\displaystyle 7\text{або }x=}$${\displaystyle 2}$;

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

ОДЗ: ${\displaystyle x\ne 0,x\ne 2}$;

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

${\displaystyle 2x^2-x-(5x^2-16x+12)=}$${\displaystyle 6x^2-24x+24}$;

${\displaystyle -3x^2+15x-12=}$${\displaystyle 6x^2-24x+24}$;

${\displaystyle -9x^2+39x-36=}$${\displaystyle 0\mid :\left(-3\right)}$;

${\displaystyle 3x^2-13x+12=}$${\displaystyle 0}$;

${\displaystyle a=}$${\displaystyle 3,b=}$${\displaystyle -13,c=}$${\displaystyle 12}$;

${\displaystyle D=}$${\displaystyle b^2-4ac=}$${\displaystyle {(-13)}^2-4\cdot 3\cdot 12=}$${\displaystyle 169-144=}$${\displaystyle 25}$;

${\displaystyle D>0}$, два корені:

${\displaystyle x_{1,2}=}$${\displaystyle \frac{-b\pm \sqrt{D}}{2a}=}$${\displaystyle \frac{13\pm \sqrt{25}}{6}=}$${\displaystyle \frac{13\pm 5}{6}}$;

${\displaystyle x_1=}$${\displaystyle \frac{13+5}{6}=}$${\displaystyle 3}$;

${\displaystyle x_2=}$${\displaystyle \frac{13-5}{6}=}$${\displaystyle \frac{4}{3}}$;

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

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

г) ${\displaystyle \frac{1}{x-1}-\frac{8x-2}{x^2+x+1}=}$${\displaystyle \frac{x-9}{(x-1)(x^2+x+1)}\mid \cdot \left(x-1\right)\left(x^2+x+1\right)}$;

ОДЗ: ${\displaystyle x\ne 1}$;

${\displaystyle x^2+x+1-(8x-2)(x-1)=}$${\displaystyle x-9}$;

${\displaystyle x^2+x+1-(8x^2-10x+2)=}$${\displaystyle x-9}$;

${\displaystyle -7x^2+11x-1=}$${\displaystyle x-9}$;

${\displaystyle -7x^2+10x+8=}$${\displaystyle 0}$;

${\displaystyle 7x^2-10x-8=}$${\displaystyle 0}$;

${\displaystyle a=}$${\displaystyle 7,b=}$${\displaystyle -10,c=}$${\displaystyle -8}$;

${\displaystyle D=}$${\displaystyle b^2-4ac=}$${\displaystyle {(-10)}^2-4\cdot 7\cdot (-8)=}$${\displaystyle 100+224=}$${\displaystyle 324}$;

${\displaystyle D>0}$, два корені:

${\displaystyle x_{1,2}=}$${\displaystyle \frac{-b\pm \sqrt{D}}{2a}=}$${\displaystyle \frac{10\pm \sqrt{324}}{14}=}$${\displaystyle \frac{10\pm 18}{14}}$;

${\displaystyle x_1=}$${\displaystyle \frac{10+18}{14}=}$${\displaystyle 2}$;

${\displaystyle x_2=}$${\displaystyle \frac{10-18}{14}=}$${\displaystyle -\frac{4}{7}}$;

Отже, ${\displaystyle x=}$${\displaystyle 2\text{або }x=}$${\displaystyle -\frac{4}{7}}$.