1097 Вправа 1097Алгебра 8 клас Бевз 2025 · § 21 · стор. 205

1097. Розв’яжи рівняння. а) 2/(x − 7) = x/(x − 2) + 10/(x² − 9x + 14); б) x/(x² − 9) − 3/(x² + 2x − 3) + (2x − 5)/(x² − 4x + 3) = 0; в) 1/(x − 1) + 1/(x² − 3x + 2) = (2x − 7)/(x² − 9x + 14).

а)

${\displaystyle x^2-9x+14}$: ${\displaystyle x_1+x_2=}$${\displaystyle 9}$;

${\displaystyle x_1{x}_2=}$${\displaystyle 14}$ → ${\displaystyle x_1=}$${\displaystyle 7,x_2=}$${\displaystyle 2}$;

${\displaystyle x^2-9x+14=}$${\displaystyle \left(x-7\right)\left(x-2\right)}$;

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

ОДЗ: ${\displaystyle x\ne 2,x\ne 7}$;

${\displaystyle 2(x-2)=}$${\displaystyle x(x-7)+10}$;

${\displaystyle 2x-4=}$${\displaystyle x^2-7x+10}$;

${\displaystyle 0=}$${\displaystyle x^2-7x+10-2x+4}$;

${\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}$;

${\displaystyle x=}$${\displaystyle 7}$ — не входить в ОДЗ (сторонній корінь);

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

Отже, коренів немає;

б)

${\displaystyle x^2+2x-3}$: ${\displaystyle x_1+x_2=}$${\displaystyle -2}$;

${\displaystyle x_1{x}_2=}$${\displaystyle -3}$ → ${\displaystyle x_1=}$${\displaystyle 1,x_2=}$${\displaystyle -3}$;

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

${\displaystyle x^2-4x+3}$: ${\displaystyle x_1+x_2=}$${\displaystyle 4}$;

${\displaystyle x_1{x}_2=}$${\displaystyle 3}$ → ${\displaystyle x_1=}$${\displaystyle 3,x_2=}$${\displaystyle 1}$;

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

${\displaystyle \frac{x}{(x-3)(x+3)}-\frac{3}{(x+3)(x-1)}+\frac{2x-5}{(x-1)(x-3)}=}$${\displaystyle 0\mid \cdot \left(x-3\right)\left(x+3\right)\left(x-1\right)}$;

ОДЗ: ${\displaystyle x\ne \pm 3,x\ne 1}$;

${\displaystyle x(x-1)-3(x-3)+(2x-5)(x+3)=}$${\displaystyle 0}$;

${\displaystyle x^2-x-3x+9+2x^2+x-15=}$${\displaystyle 0}$;

${\displaystyle 3x^2-3x-6=}$${\displaystyle 0\mid :3}$;

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

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

${\displaystyle D=}$${\displaystyle b^2-4ac=}$${\displaystyle {(-1)}^2-4\cdot 1\cdot (-2)=}$${\displaystyle 1+8=}$${\displaystyle 9}$;

${\displaystyle D>0}$, два корені:

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

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

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

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

в)

${\displaystyle x^2-3x+2}$: ${\displaystyle x_1+x_2=}$${\displaystyle 3}$;

${\displaystyle x_1{x}_2=}$${\displaystyle 2}$ → ${\displaystyle x_1=}$${\displaystyle 2,x_2=}$${\displaystyle 1}$;

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

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

ОДЗ: ${\displaystyle x\ne 1,x\ne 2,x\ne 7}$;

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

${\displaystyle x^2-8x+7=}$${\displaystyle 2x^2-9x+7}$;

${\displaystyle 0=}$${\displaystyle 2x^2-9x+7-x^2+8x-7}$;

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

${\displaystyle a=}$${\displaystyle 1,b=}$${\displaystyle -1,c=}$${\displaystyle 0}$;

${\displaystyle D=}$${\displaystyle b^2-4ac=}$${\displaystyle {(-1)}^2-4\cdot 1\cdot 0=}$${\displaystyle 1-0=}$${\displaystyle 1}$;

${\displaystyle D>0}$, два корені:

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

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

${\displaystyle x_2=}$${\displaystyle \frac{1-1}{2}=}$${\displaystyle 0}$;

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

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