970 Вправа 970Алгебра 8 клас Бевз 2025 · § 19 · стор. 184

970. Розв’яжи рівняння. а) x + 3 = (x + 3)/x; б) y = 18/(y − 3); в) (z + 2)/z = (5z + 1)/(z + 1); г) (5 + 3m)/(5 − m) = (m + 3)/(m + 1); ґ) (x² − 5x + 6)/(x − 3) = 0; д) (2x² + 5x + 2)/(x² − 4) = 0.

а) ${\displaystyle x+3=}$${\displaystyle \frac{x+3}{x}\mid \cdot x}$;

ОДЗ: ${\displaystyle x\ne 0}$;

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

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

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

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

${\displaystyle D=}$${\displaystyle b^2-4ac=}$${\displaystyle 2^2-4\cdot 1\cdot (-3)=}$${\displaystyle 4+12=}$${\displaystyle 16}$;

${\displaystyle D>0}$, два корені:

${\displaystyle x_{1,2}=}$${\displaystyle \frac{-b\pm \sqrt{D}}{2a}=}$${\displaystyle \frac{-2\pm \sqrt{16}}{2}=}$${\displaystyle \frac{-2\pm 4}{2}}$;

${\displaystyle x_1=}$${\displaystyle \frac{-2+4}{2}=}$${\displaystyle 1}$;

${\displaystyle x_2=}$${\displaystyle \frac{-2-4}{2}=}$${\displaystyle -3}$;

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

б) ${\displaystyle y=}$${\displaystyle \frac{18}{y-3}\mid \cdot \left(y-3\right)}$;

ОДЗ: ${\displaystyle y\ne 3}$;

${\displaystyle y(y-3)=}$${\displaystyle 18}$;

${\displaystyle y^2-3y-18=}$${\displaystyle 0}$;

${\displaystyle a=}$${\displaystyle 1,b=}$${\displaystyle -3,c=}$${\displaystyle -18}$;

${\displaystyle D=}$${\displaystyle b^2-4ac=}$${\displaystyle {(-3)}^2-4\cdot 1\cdot (-18)=}$${\displaystyle 9+72=}$${\displaystyle 81}$;

${\displaystyle D>0}$, два корені:

${\displaystyle y_{1,2}=}$${\displaystyle \frac{-b\pm \sqrt{D}}{2a}=}$${\displaystyle \frac{3\pm \sqrt{81}}{2}=}$${\displaystyle \frac{3\pm 9}{2}}$;

${\displaystyle y_1=}$${\displaystyle \frac{3+9}{2}=}$${\displaystyle 6}$;

${\displaystyle y_2=}$${\displaystyle \frac{3-9}{2}=}$${\displaystyle -3}$;

Отже, ${\displaystyle y=}$${\displaystyle 6\text{або }y=}$${\displaystyle -3}$;

в) ${\displaystyle \frac{z+2}{z}=}$${\displaystyle \frac{5z+1}{z+1}\mid \cdot \left(z\left(z+1\right)\right)}$;

ОДЗ: ${\displaystyle z\ne 0,z\ne -1}$;

${\displaystyle (z+2)(z+1)=}$${\displaystyle z(5z+1)}$;

${\displaystyle z^2+3z+2=}$${\displaystyle 5z^2+z}$;

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

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

${\displaystyle 2z^2-z-1=}$${\displaystyle 0}$;

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

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

${\displaystyle D>0}$, два корені:

${\displaystyle z_{1,2}=}$${\displaystyle \frac{-b\pm \sqrt{D}}{2a}=}$${\displaystyle \frac{1\pm \sqrt{9}}{4}=}$${\displaystyle \frac{1\pm 3}{4}}$;

${\displaystyle z_1=}$${\displaystyle \frac{1+3}{4}=}$${\displaystyle 1}$;

${\displaystyle z_2=}$${\displaystyle \frac{1-3}{4}=}$${\displaystyle -0,5}$;

Отже, ${\displaystyle z=}$${\displaystyle 1\text{або }z=}$${\displaystyle -0,5}$;

г) ${\displaystyle \frac{5+3m}{5-m}=}$${\displaystyle \frac{m+3}{m+1}\mid \cdot \left(5-m\right)\left(m+1\right)}$;

ОДЗ: ${\displaystyle m\ne 5,m\ne -1}$;

${\displaystyle (5+3m)(m+1)=}$${\displaystyle (m+3)(5-m)}$;

${\displaystyle 3m^2+8m+5=}$${\displaystyle -m^2+2m+15}$;

${\displaystyle 3m^2+8m+5+m^2-2m-15=}$${\displaystyle 0}$;

${\displaystyle 4m^2+6m-10=}$${\displaystyle 0\mid :2}$;

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

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

${\displaystyle D=}$${\displaystyle b^2-4ac=}$${\displaystyle 3^2-4\cdot 2\cdot (-5)=}$${\displaystyle 9+40=}$${\displaystyle 49}$;

${\displaystyle D>0}$, два корені:

${\displaystyle m_{1,2}=}$${\displaystyle \frac{-b\pm \sqrt{D}}{2a}=}$${\displaystyle \frac{-3\pm \sqrt{49}}{4}=}$${\displaystyle \frac{-3\pm 7}{4}}$;

${\displaystyle m_1=}$${\displaystyle \frac{-3+7}{4}=}$${\displaystyle 1}$;

${\displaystyle m_2=}$${\displaystyle \frac{-3-7}{4}=}$${\displaystyle -2,5}$;

Отже, ${\displaystyle m=}$${\displaystyle 1\text{або }m=}$${\displaystyle -2,5}$;

ґ) ${\displaystyle \frac{x^2-5x+6}{x-3}=}$${\displaystyle 0}$;

ОДЗ: ${\displaystyle x\ne 3}$;

дріб дорівнює нулю, якщо чисельник дорівнює нулю, а знаменник — ні:

${\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 1}$;

${\displaystyle D>0}$, два корені:

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

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

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

${\displaystyle x=}$${\displaystyle 3}$ — не входить в ОДЗ (сторонній корінь);

Отже, ${\displaystyle x=}$${\displaystyle 2}$;

д) ${\displaystyle \frac{2x^2+5x+2}{x^2-4}=}$${\displaystyle 0}$;

ОДЗ: ${\displaystyle x\ne \pm 2}$;

дріб дорівнює нулю, якщо чисельник дорівнює нулю, а знаменник — ні:

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

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

${\displaystyle D=}$${\displaystyle b^2-4ac=}$${\displaystyle 5^2-4\cdot 2\cdot 2=}$${\displaystyle 25-16=}$${\displaystyle 9}$;

${\displaystyle D>0}$, два корені:

${\displaystyle x_{1,2}=}$${\displaystyle \frac{-b\pm \sqrt{D}}{2a}=}$${\displaystyle \frac{-5\pm \sqrt{9}}{4}=}$${\displaystyle \frac{-5\pm 3}{4}}$;

${\displaystyle x_1=}$${\displaystyle \frac{-5+3}{4}=}$${\displaystyle -0,5}$;

${\displaystyle x_2=}$${\displaystyle \frac{-5-3}{4}=}$${\displaystyle -2}$;

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

Отже, ${\displaystyle x=}$${\displaystyle -0,5}$.