47 Вправа 47 (повторення)Алгебра 8 клас Тарасенкова 2025 · Повторення · стор. 321

47. Розв’яжіть рівняння:

1) ${\displaystyle 3-\frac{14}{x}=}$${\displaystyle \frac{2}{x-5}}$;

ОДЗ: ${\displaystyle x\ne 0}$; ${\displaystyle x-5\ne 0}$;

${\displaystyle x\ne 5}$;

${\displaystyle 3^{\mathrm{\setminus }x\left(x-5\right)}-\frac{{14}^{\mathrm{\setminus }x-5}}{x}-\frac{2^{\mathrm{\setminus }x}}{x-5}=}$${\displaystyle 0}$;

${\displaystyle \frac{3x(x-5)-14(x-5)-2x}{x(x-5)}=}$${\displaystyle 0}$;

${\displaystyle 3x(x-5)-14(x-5)-2x=}$${\displaystyle 0}$;

${\displaystyle 3x^2-15x-14x+70-2x=}$${\displaystyle 0}$;

${\displaystyle 3x^2-31x+70=}$${\displaystyle 0}$;

${\displaystyle D=}$${\displaystyle {(-31)}^2-4\cdot 3\cdot 70=}$${\displaystyle 961-840=}$${\displaystyle 121}$;

${\displaystyle x_{1,2}=}$${\displaystyle \frac{31\pm \sqrt{121}}{2\cdot 3}=}$${\displaystyle \frac{31\pm 11}{6}}$;

${\displaystyle x_1=}$${\displaystyle \frac{31+11}{6}=}$${\displaystyle 7}$;

${\displaystyle x_2=}$${\displaystyle \frac{31-11}{6}=}$${\displaystyle \frac{20}{6}=}$${\displaystyle 3\frac{1}{3}}$;

обидва корені задовольняють ОДЗ;

2) ${\displaystyle \frac{10}{x-3}-1=}$${\displaystyle \frac{8}{x}}$;

ОДЗ: ${\displaystyle x-3\ne 0}$; ${\displaystyle x\ne 0}$;

${\displaystyle x\ne 3}$;

${\displaystyle \frac{{10}^{\mathrm{\setminus }x}}{x-3}-1^{\mathrm{\setminus }x\left(x-3\right)}-\frac{8^{\mathrm{\setminus }x-3}}{x}=}$${\displaystyle 0}$;

${\displaystyle \frac{10x-x(x-3)-8(x-3)}{x(x-3)}=}$${\displaystyle 0}$;

${\displaystyle 10x-x(x-3)-8(x-3)=}$${\displaystyle 0}$;

${\displaystyle 10x-x^2+3x-8x+24=}$${\displaystyle 0}$;

${\displaystyle -x^2+5x+24=}$${\displaystyle 0}$ | ${\displaystyle \cdot \left(-1\right)}$

${\displaystyle x^2-5x-24=}$${\displaystyle 0}$;

${\displaystyle D=}$${\displaystyle {(-5)}^2-4\cdot 1\cdot (-24)=}$${\displaystyle 25+96=}$${\displaystyle 121}$;

${\displaystyle x_{1,2}=}$${\displaystyle \frac{5\pm \sqrt{121}}{2\cdot 1}=}$${\displaystyle \frac{5\pm 11}{2}}$;

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

${\displaystyle x_2=}$${\displaystyle \frac{5-11}{2}=}$${\displaystyle -3}$;

обидва корені задовольняють ОДЗ;

47. Розв’яжіть рівняння (продовження):

3) ${\displaystyle \frac{1}{x+3}+\frac{1}{x}=}$${\displaystyle \frac{3}{20}}$;

ОДЗ: ${\displaystyle x+3\ne 0}$; ${\displaystyle x\ne 0}$;

${\displaystyle x\ne -3}$;

${\displaystyle \frac{1^{\mathrm{\setminus }20x}}{x+3}+\frac{1^{\mathrm{\setminus }20\left(x+3\right)}}{x}-\frac{3^{\mathrm{\setminus }x\left(x+3\right)}}{20}=}$${\displaystyle 0}$;

${\displaystyle \frac{20x+20(x+3)-3x(x+3)}{20x(x+3)}=}$${\displaystyle 0}$;

${\displaystyle 20x+20(x+3)-3x(x+3)=}$${\displaystyle 0}$;

${\displaystyle 20x+20x+60-3x^2-9x=}$${\displaystyle 0}$;

${\displaystyle -3x^2+31x+60=}$${\displaystyle 0}$ | ${\displaystyle \cdot \left(-1\right)}$

${\displaystyle 3x^2-31x-60=}$${\displaystyle 0}$;

${\displaystyle D=}$${\displaystyle {(-31)}^2-4\cdot 3\cdot (-60)=}$${\displaystyle 961+720=}$${\displaystyle 1681}$;

${\displaystyle x_{1,2}=}$${\displaystyle \frac{31\pm \sqrt{1681}}{2\cdot 3}=}$${\displaystyle \frac{31\pm 41}{6}}$;

${\displaystyle x_1=}$${\displaystyle \frac{31+41}{6}=}$${\displaystyle 12}$;

${\displaystyle x_2=}$${\displaystyle \frac{31-41}{6}=}$${\displaystyle \frac{-10}{6}=}$${\displaystyle -1\frac{2}{3}}$;

обидва корені задовольняють ОДЗ;

4) ${\displaystyle -\frac{1}{x-3}=}$${\displaystyle \frac{1}{x}+1\frac{1}{4}}$;

ОДЗ: ${\displaystyle x-3\ne 0}$; ${\displaystyle x\ne 0}$;

${\displaystyle x\ne 3}$;

${\displaystyle -\frac{1}{x-3}=}$${\displaystyle \frac{1}{x}+\frac{5}{4}}$;

${\displaystyle -\frac{1^{\mathrm{\setminus }4x}}{x-3}-\frac{1^{\mathrm{\setminus }4\left(x-3\right)}}{x}-\frac{5^{\mathrm{\setminus }x\left(x-3\right)}}{4}=}$${\displaystyle 0}$;

${\displaystyle \frac{-4x-4(x-3)-5x(x-3)}{4x(x-3)}=}$${\displaystyle 0}$;

${\displaystyle -4x-4(x-3)-5x(x-3)=}$${\displaystyle 0}$;

${\displaystyle -4x-4x+12-5x^2+15x=}$${\displaystyle 0}$;

${\displaystyle -5x^2+7x+12=}$${\displaystyle 0}$ | ${\displaystyle \cdot \left(-1\right)}$

${\displaystyle 5x^2-7x-12=}$${\displaystyle 0}$;

${\displaystyle D=}$${\displaystyle {(-7)}^2-4\cdot 5\cdot (-12)=}$${\displaystyle 49+240=}$${\displaystyle 289}$;

${\displaystyle x_{1,2}=}$${\displaystyle \frac{7\pm \sqrt{289}}{2\cdot 5}=}$${\displaystyle \frac{7\pm 17}{10}}$;

${\displaystyle x_1=}$${\displaystyle \frac{7+17}{10}=}$${\displaystyle \frac{24}{10}=}$${\displaystyle 2,4}$;

${\displaystyle x_2=}$${\displaystyle \frac{7-17}{10}=}$${\displaystyle -1}$;

обидва корені задовольняють ОДЗ.