988 Вправа 988Алгебра 8 клас Бевз 2025 · § 19 · стор. 185

988. Розв’яжи рівняння. а) (x² − 4x)/(x + 1) = 5/(x + 1); б) (x² − x)/(x − 4) = (5x − 8)/(x − 4); в) (1 + 3x)/(2 + x) + (x − 1)/(2 − x) = 1; г) (2y − 2)/(y + 3) − (3y − 5)/(3 − y) = 6; ґ) 5/(z − 2) + 4/(z − 3) = 5/z; д) (3x + 1)/(x − 5) + (6x − 2)/(x + 5) = (5x² + 3)/(x² − 25).

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

ОДЗ: ${\displaystyle x\ne -1}$;

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

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

${\displaystyle a=}$${\displaystyle 1,b=}$${\displaystyle -4,c=}$${\displaystyle -5}$;

${\displaystyle D=}$${\displaystyle b^2-4ac=}$${\displaystyle {(-4)}^2-4\cdot 1\cdot (-5)=}$${\displaystyle 16+20=}$${\displaystyle 36}$;

${\displaystyle D>0}$, два корені:

${\displaystyle x_{1,2}=}$${\displaystyle \frac{-b\pm \sqrt{D}}{2a}=}$${\displaystyle \frac{4\pm \sqrt{36}}{2}=}$${\displaystyle \frac{4\pm 6}{2}}$;

${\displaystyle x_1=}$${\displaystyle \frac{4+6}{2}=}$${\displaystyle 5}$;

${\displaystyle x_2=}$${\displaystyle \frac{4-6}{2}=}$${\displaystyle -1}$;

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

Отже, ${\displaystyle x=}$${\displaystyle 5}$;

б) ${\displaystyle \frac{x^2-x}{x-4}=}$${\displaystyle \frac{5x-8}{x-4}\mid \cdot \left(x-4\right)}$;

ОДЗ: ${\displaystyle x\ne 4}$;

${\displaystyle x^2-x=}$${\displaystyle 5x-8}$;

${\displaystyle x^2-x-5x+8=}$${\displaystyle 0}$;

${\displaystyle x^2-6x+8=}$${\displaystyle 0}$;

${\displaystyle a=}$${\displaystyle 1,b=}$${\displaystyle -6,c=}$${\displaystyle 8}$;

${\displaystyle D=}$${\displaystyle b^2-4ac=}$${\displaystyle {(-6)}^2-4\cdot 1\cdot 8=}$${\displaystyle 36-32=}$${\displaystyle 4}$;

${\displaystyle D>0}$, два корені:

${\displaystyle x_{1,2}=}$${\displaystyle \frac{-b\pm \sqrt{D}}{2a}=}$${\displaystyle \frac{6\pm \sqrt{4}}{2}=}$${\displaystyle \frac{6\pm 2}{2}}$;

${\displaystyle x_1=}$${\displaystyle \frac{6+2}{2}=}$${\displaystyle 4}$;

${\displaystyle x_2=}$${\displaystyle \frac{6-2}{2}=}$${\displaystyle 2}$;

${\displaystyle x=}$${\displaystyle 4}$ — не входить в ОДЗ (сторонній корінь);

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

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

ОДЗ: ${\displaystyle x\ne \pm 2}$;

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

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

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

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

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

${\displaystyle a=}$${\displaystyle 1,b=}$${\displaystyle -6,c=}$${\displaystyle 4}$;

${\displaystyle D=}$${\displaystyle b^2-4ac=}$${\displaystyle {(-6)}^2-4\cdot 1\cdot 4=}$${\displaystyle 36-16=}$${\displaystyle 20}$;

${\displaystyle D>0}$, два корені:

${\displaystyle x_{1,2}=}$${\displaystyle \frac{-b\pm \sqrt{D}}{2a}=}$${\displaystyle \frac{6\pm \sqrt{20}}{2}=}$${\displaystyle \frac{6\pm 2\sqrt{5}}{2}=}$${\displaystyle 3\pm \sqrt{5}}$;

Обидва корені входять в ОДЗ: ${\displaystyle x=}$${\displaystyle 3\pm \sqrt{5}}$;

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

ОДЗ: ${\displaystyle y\ne -3,y\ne 3}$;

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

${\displaystyle -2y^2+8y-6-(3y^2+4y-15)=}$${\displaystyle 6(9-y^2)}$;

${\displaystyle -5y^2+4y+9=}$${\displaystyle 54-6y^2}$;

${\displaystyle -5y^2+4y+9-54+6y^2=}$${\displaystyle 0}$;

${\displaystyle y^2+4y-45=}$${\displaystyle 0}$;

${\displaystyle a=}$${\displaystyle 1,b=}$${\displaystyle 4,c=}$${\displaystyle -45}$;

${\displaystyle D=}$${\displaystyle b^2-4ac=}$${\displaystyle 4^2-4\cdot 1\cdot (-45)=}$${\displaystyle 16+180=}$${\displaystyle 196}$;

${\displaystyle D>0}$, два корені:

${\displaystyle y_{1,2}=}$${\displaystyle \frac{-b\pm \sqrt{D}}{2a}=}$${\displaystyle \frac{-4\pm \sqrt{196}}{2}=}$${\displaystyle \frac{-4\pm 14}{2}}$;

${\displaystyle y_1=}$${\displaystyle \frac{-4+14}{2}=}$${\displaystyle 5}$;

${\displaystyle y_2=}$${\displaystyle \frac{-4-14}{2}=}$${\displaystyle -9}$;

Отже, ${\displaystyle y=}$${\displaystyle 5\text{або }y=}$${\displaystyle -9}$;

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

ОДЗ: ${\displaystyle z\ne 0,z\ne 2,z\ne 3}$;

${\displaystyle 5z(z-3)+4z(z-2)=}$${\displaystyle 5(z-2)(z-3)}$;

${\displaystyle 5z^2-15z+4z^2-8z=}$${\displaystyle 5z^2-25z+30}$;

${\displaystyle 9z^2-23z-5z^2+25z-30=}$${\displaystyle 0}$;

${\displaystyle 4z^2+2z-30=}$${\displaystyle 0\mid :2}$;

${\displaystyle 2z^2+z-15=}$${\displaystyle 0}$;

${\displaystyle a=}$${\displaystyle 2,b=}$${\displaystyle 1,c=}$${\displaystyle -15}$;

${\displaystyle D=}$${\displaystyle b^2-4ac=}$${\displaystyle 1^2-4\cdot 2\cdot (-15)=}$${\displaystyle 1+120=}$${\displaystyle 121}$;

${\displaystyle D>0}$, два корені:

${\displaystyle z_{1,2}=}$${\displaystyle \frac{-b\pm \sqrt{D}}{2a}=}$${\displaystyle \frac{-1\pm \sqrt{121}}{4}=}$${\displaystyle \frac{-1\pm 11}{4}}$;

${\displaystyle z_1=}$${\displaystyle \frac{-1+11}{4}=}$${\displaystyle 2,5}$;

${\displaystyle z_2=}$${\displaystyle \frac{-1-11}{4}=}$${\displaystyle -3}$;

Отже, ${\displaystyle z=}$${\displaystyle 2,5\text{або }z=}$${\displaystyle -3}$;

д) ${\displaystyle \frac{3x+1}{x-5}+\frac{6x-2}{x+5}=}$${\displaystyle \frac{5x^2+3}{x^2-25}\mid \cdot \left(x-5\right)\left(x+5\right)}$;

ОДЗ: ${\displaystyle x\ne \pm 5}$;

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

${\displaystyle 3x^2+16x+5+6x^2-32x+10=}$${\displaystyle 5x^2+3}$;

${\displaystyle 9x^2-16x+15-5x^2-3=}$${\displaystyle 0}$;

${\displaystyle 4x^2-16x+12=}$${\displaystyle 0\mid :4}$;

${\displaystyle x^2-4x+3=}$${\displaystyle 0}$;

${\displaystyle a=}$${\displaystyle 1,b=}$${\displaystyle -4,c=}$${\displaystyle 3}$;

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

${\displaystyle D>0}$, два корені:

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

${\displaystyle x_1=}$${\displaystyle \frac{4+2}{2}=}$${\displaystyle 3}$;

${\displaystyle x_2=}$${\displaystyle \frac{4-2}{2}=}$${\displaystyle 1}$;

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