1001 Вправа 1001Алгебра 8 клас Бевз 2025 · § 19 · стор. 187

1001. Розв’яжи рівняння. а) (2x + 1)/(x + 2) + (x + 2)/(2x + 1) = 10/3; б) 5x/(x − 4) − (x − 4)/x = −4.

а) ${\displaystyle \frac{2x+1}{x+2}+\frac{x+2}{2x+1}=}$${\displaystyle \frac{10}{3}}$;

заміна ${\displaystyle \frac{2x+1}{x+2}=}$${\displaystyle t}$, ${\displaystyle t\ne 0}$ (${\displaystyle x\ne -2}$, ${\displaystyle x\ne -0,5}$):

${\displaystyle t+\frac{1}{t}=}$${\displaystyle \frac{10}{3}}$;

${\displaystyle 3t^2+3=}$${\displaystyle 10t}$;

${\displaystyle 3t^2-10t+3=}$${\displaystyle 0}$;

${\displaystyle a=}$${\displaystyle 3,b=}$${\displaystyle -10,c=}$${\displaystyle 3}$;

${\displaystyle D=}$${\displaystyle b^2-4ac=}$${\displaystyle {(-10)}^2-4\cdot 3\cdot 3=}$${\displaystyle 100-36=}$${\displaystyle 64}$;

${\displaystyle D>0}$, два корені:

${\displaystyle t_{1,2}=}$${\displaystyle \frac{-b\pm \sqrt{D}}{2a}=}$${\displaystyle \frac{10\pm \sqrt{64}}{6}=}$${\displaystyle \frac{10\pm 8}{6}}$;

${\displaystyle t_1=}$${\displaystyle \frac{10+8}{6}=}$${\displaystyle 3}$;

${\displaystyle t_2=}$${\displaystyle \frac{10-8}{6}=}$${\displaystyle \frac{1}{3}}$;

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

${\displaystyle 2x+1=}$${\displaystyle 3x+6}$;

${\displaystyle x=}$${\displaystyle -5}$;

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

${\displaystyle 6x+3=}$${\displaystyle x+2}$;

${\displaystyle 5x=}$${\displaystyle -1}$;

${\displaystyle x=}$${\displaystyle -0,2}$;

б) ${\displaystyle \frac{5x}{x-4}-\frac{x-4}{x}=}$${\displaystyle -4}$;

заміна ${\displaystyle \frac{x}{x-4}=}$${\displaystyle t}$, ${\displaystyle t\ne 0}$ (${\displaystyle x\ne 0}$, ${\displaystyle x\ne 4}$):

${\displaystyle 5t-\frac{1}{t}=}$${\displaystyle -4}$;

${\displaystyle 5t^2-1=}$${\displaystyle -4t}$;

${\displaystyle 5t^2+4t-1=}$${\displaystyle 0}$;

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

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

${\displaystyle D>0}$, два корені:

${\displaystyle t_{1,2}=}$${\displaystyle \frac{-b\pm \sqrt{D}}{2a}=}$${\displaystyle \frac{-4\pm \sqrt{36}}{10}=}$${\displaystyle \frac{-4\pm 6}{10}}$;

${\displaystyle t_1=}$${\displaystyle \frac{-4+6}{10}=}$${\displaystyle 0,2}$;

${\displaystyle t_2=}$${\displaystyle \frac{-4-6}{10}=}$${\displaystyle -1}$;

${\displaystyle \frac{x}{x-4}=}$${\displaystyle 0,2}$;

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

${\displaystyle x=}$${\displaystyle -1}$;

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

${\displaystyle x=}$${\displaystyle -x+4}$;

${\displaystyle x=}$${\displaystyle 2}$.