1096 Вправа 1096Алгебра 8 клас Бевз 2025 · § 21 · стор. 205

1096. Розв’яжи рівняння. а) x/(x − 2) + 6/(x² − 7x + 10) = 2/(x − 5); б) 2/(x² + 3x + 2) − x/(x² − 1) + (5x + 4)/(x² + x − 2) = 0; в) 1/(x − 1) + 3/(x² + 9x + 18) = (2x + 5)/(x² + 5x − 6).

Розкладаємо знаменники на множники, знаходимо ОДЗ і множимо на спільний знаменник.

а)

${\displaystyle x^2-7x+10}$: ${\displaystyle x_1+x_2=}$${\displaystyle 7}$;

${\displaystyle x_1{x}_2=}$${\displaystyle 10}$ → ${\displaystyle x_1=}$${\displaystyle 5,x_2=}$${\displaystyle 2}$;

${\displaystyle x^2-7x+10=}$${\displaystyle \left(x-5\right)\left(x-2\right)}$;

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

ОДЗ: ${\displaystyle x\ne 2,x\ne 5}$;

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

${\displaystyle x^2-5x+6-2x+4=}$${\displaystyle 0}$;

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

${\displaystyle a=}$${\displaystyle 1,b=}$${\displaystyle -7,c=}$${\displaystyle 10}$;

${\displaystyle D=}$${\displaystyle b^2-4ac=}$${\displaystyle {(-7)}^2-4\cdot 1\cdot 10=}$${\displaystyle 49-40=}$${\displaystyle 9}$;

${\displaystyle D>0}$, два корені:

${\displaystyle x_{1,2}=}$${\displaystyle \frac{-b\pm \sqrt{D}}{2a}=}$${\displaystyle \frac{7\pm \sqrt{9}}{2}=}$${\displaystyle \frac{7\pm 3}{2}}$;

${\displaystyle x_1=}$${\displaystyle \frac{7+3}{2}=}$${\displaystyle 5}$;

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

${\displaystyle x=}$${\displaystyle 5}$ — не входить в ОДЗ (сторонній корінь);

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

Отже, коренів немає;

б)

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

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

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

${\displaystyle x^2+x-2}$: ${\displaystyle x_1+x_2=}$${\displaystyle -1}$;

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

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

${\displaystyle \frac{2}{(x+1)(x+2)}-\frac{x}{(x-1)(x+1)}+\frac{5x+4}{(x+2)(x-1)}=}$${\displaystyle 0\mid \cdot \left(x+1\right)\left(x+2\right)\left(x-1\right)}$;

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

${\displaystyle 2(x-1)-x(x+2)+(5x+4)(x+1)=}$${\displaystyle 0}$;

${\displaystyle 2x-2-x^2-2x+5x^2+9x+4=}$${\displaystyle 0}$;

${\displaystyle 4x^2+9x+2=}$${\displaystyle 0}$;

${\displaystyle a=}$${\displaystyle 4,b=}$${\displaystyle 9,c=}$${\displaystyle 2}$;

${\displaystyle D=}$${\displaystyle b^2-4ac=}$${\displaystyle 9^2-4\cdot 4\cdot 2=}$${\displaystyle 81-32=}$${\displaystyle 49}$;

${\displaystyle D>0}$, два корені:

${\displaystyle x_{1,2}=}$${\displaystyle \frac{-b\pm \sqrt{D}}{2a}=}$${\displaystyle \frac{-9\pm \sqrt{49}}{8}=}$${\displaystyle \frac{-9\pm 7}{8}}$;

${\displaystyle x_1=}$${\displaystyle \frac{-9+7}{8}=}$${\displaystyle -0,25}$;

${\displaystyle x_2=}$${\displaystyle \frac{-9-7}{8}=}$${\displaystyle -2}$;

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

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

в)

${\displaystyle x^2+9x+18}$: ${\displaystyle x_1+x_2=}$${\displaystyle -9}$;

${\displaystyle x_1{x}_2=}$${\displaystyle 18}$ → ${\displaystyle x_1=}$${\displaystyle -3,x_2=}$${\displaystyle -6}$;

${\displaystyle x^2+9x+18=}$${\displaystyle \left(x+3\right)\left(x+6\right)}$;

${\displaystyle x^2+5x-6}$: ${\displaystyle x_1+x_2=}$${\displaystyle -5}$;

${\displaystyle x_1{x}_2=}$${\displaystyle -6}$ → ${\displaystyle x_1=}$${\displaystyle 1,x_2=}$${\displaystyle -6}$;

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

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

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

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

${\displaystyle x^2+12x+15=}$${\displaystyle 2x^2+11x+15}$;

${\displaystyle 0=}$${\displaystyle 2x^2+11x+15-x^2-12x-15}$;

${\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}$.