840° Вправа 840Алгебра 8 клас Тарасенкова 2025 · § 21 · стор. 254

840. Розв’яжіть рівняння:

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

${\displaystyle x^2=}$${\displaystyle 5x-6}$, ${\displaystyle x\ne 3}$;

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

${\displaystyle D=}$${\displaystyle {(-5)}^2-4\cdot 1\cdot 6=}$${\displaystyle 25-24=}$${\displaystyle 1}$;

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

${\displaystyle x_2=}$${\displaystyle \frac{5+1}{2}=}$${\displaystyle 3}$ — не задовольняє умову ${\displaystyle x\ne 3}$;

2) ${\displaystyle \frac{x^2-15}{5-x}=}$${\displaystyle \frac{2x}{5-x}}$;

${\displaystyle x^2-15=}$${\displaystyle 2x}$, ${\displaystyle x\ne 5}$;

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

${\displaystyle D=}$${\displaystyle {(-2)}^2-4\cdot 1\cdot (-15)=}$${\displaystyle 4+60=}$${\displaystyle 64}$;

${\displaystyle x_1=}$${\displaystyle \frac{2-8}{2}=}$${\displaystyle -3}$;

${\displaystyle x_2=}$${\displaystyle \frac{2+8}{2}=}$${\displaystyle 5}$ — не задовольняє умову ${\displaystyle x\ne 5}$.

3) ${\displaystyle \frac{8}{x-1}=}$${\displaystyle \frac{6x+x^2}{1-x}}$;

${\displaystyle \frac{8}{x-1}=}$${\displaystyle \frac{-(6x+x^2)}{x-1}}$;

${\displaystyle 8=}$${\displaystyle -6x-x^2}$, ${\displaystyle x\ne 1}$;

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

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

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

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

4) ${\displaystyle \frac{x}{5-x}=}$${\displaystyle \frac{x^2-30}{x-5}}$;

${\displaystyle \frac{-x}{x-5}=}$${\displaystyle \frac{x^2-30}{x-5}}$;

${\displaystyle -x=}$${\displaystyle x^2-30}$, ${\displaystyle x\ne 5}$;

${\displaystyle x^2+x-30=}$${\displaystyle 0}$;

${\displaystyle D=}$${\displaystyle 1^2-4\cdot 1\cdot (-30)=}$${\displaystyle 1+120=}$${\displaystyle 121}$;

${\displaystyle x_1=}$${\displaystyle \frac{-1-11}{2}=}$${\displaystyle -6}$;

${\displaystyle x_2=}$${\displaystyle \frac{-1+11}{2}=}$${\displaystyle 5}$ — не задовольняє умову ${\displaystyle x\ne 5}$.

5) ${\displaystyle \frac{3x}{x-6}=}$${\displaystyle \frac{x^2-18}{6-x}}$;

${\displaystyle \frac{3x}{x-6}=}$${\displaystyle \frac{-(x^2-18)}{x-6}}$;

${\displaystyle 3x=}$${\displaystyle -x^2+18}$, ${\displaystyle x\ne 6}$;

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

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

${\displaystyle x_1=}$${\displaystyle \frac{-3-9}{2}=}$${\displaystyle -6}$;

${\displaystyle x_2=}$${\displaystyle \frac{-3+9}{2}=}$${\displaystyle 3}$;

6) ${\displaystyle \frac{x^2-18}{x-2}=}$${\displaystyle \frac{7x}{2-x}}$;

${\displaystyle \frac{x^2-18}{x-2}=}$${\displaystyle \frac{-7x}{x-2}}$;

${\displaystyle x^2-18=}$${\displaystyle -7x}$, ${\displaystyle x\ne 2}$;

${\displaystyle x^2+7x-18=}$${\displaystyle 0}$;

${\displaystyle D=}$${\displaystyle 7^2-4\cdot 1\cdot (-18)=}$${\displaystyle 49+72=}$${\displaystyle 121}$;

${\displaystyle x_1=}$${\displaystyle \frac{-7-11}{2}=}$${\displaystyle -9}$;

${\displaystyle x_2=}$${\displaystyle \frac{-7+11}{2}=}$${\displaystyle 2}$ — не задовольняє умову ${\displaystyle x\ne 2}$.

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

${\displaystyle \frac{2\cdot x^2}{2(x-2)}=}$${\displaystyle \frac{5x-2}{2(x-2)}}$;

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

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

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

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

${\displaystyle x_2=}$${\displaystyle \frac{5+3}{4}=}$${\displaystyle 2}$ — не задовольняє умову ${\displaystyle x\ne 2}$;

8) ${\displaystyle \frac{5x^2+3}{8x+8}=}$${\displaystyle \frac{x}{x+1}}$;

${\displaystyle \frac{5x^2+3}{8(x+1)}=}$${\displaystyle \frac{8x}{8(x+1)}}$;

${\displaystyle 5x^2+3=}$${\displaystyle 8x}$, ${\displaystyle x\ne -1}$;

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

${\displaystyle D=}$${\displaystyle {(-8)}^2-4\cdot 5\cdot 3=}$${\displaystyle 64-60=}$${\displaystyle 4}$;

${\displaystyle x_1=}$${\displaystyle \frac{8-2}{10}=}$${\displaystyle 0,6}$;

${\displaystyle x_2=}$${\displaystyle \frac{8+2}{10}=}$${\displaystyle 1}$.