1000 Вправа 1000Алгебра 8 клас Бевз 2025 · § 19 · стор. 187

1000. Розв’яжи рівняння. а) (x + 3)/(x − 1) − (x − 1)/(x + 3) = 3/2; б) x/(x + 2) − 3(x + 2)/x = 2.

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

заміна ${\displaystyle \frac{x+3}{x-1}=}$${\displaystyle t}$, ${\displaystyle t\ne 0}$ (${\displaystyle x\ne 1}$, ${\displaystyle x\ne -3}$):

${\displaystyle t-\frac{1}{t}=}$${\displaystyle \frac{3}{2}}$;

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

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

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

${\displaystyle D=}$${\displaystyle b^2-4ac=}$${\displaystyle {(-3)}^2-4\cdot 2\cdot (-2)=}$${\displaystyle 9+16=}$${\displaystyle 25}$;

${\displaystyle D>0}$, два корені:

${\displaystyle t_{1,2}=}$${\displaystyle \frac{-b\pm \sqrt{D}}{2a}=}$${\displaystyle \frac{3\pm \sqrt{25}}{4}=}$${\displaystyle \frac{3\pm 5}{4}}$;

${\displaystyle t_1=}$${\displaystyle \frac{3+5}{4}=}$${\displaystyle 2}$;

${\displaystyle t_2=}$${\displaystyle \frac{3-5}{4}=}$${\displaystyle -0,5}$;

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

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

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

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

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

${\displaystyle 3x=}$${\displaystyle -5}$;

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

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

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

${\displaystyle t-\frac{3}{t}=}$${\displaystyle 2}$;

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

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

${\displaystyle t_1=}$${\displaystyle \frac{2+4}{2}=}$${\displaystyle 3}$;

${\displaystyle t_2=}$${\displaystyle \frac{2-4}{2}=}$${\displaystyle -1}$;

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

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

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

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

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

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