1390 Вправа 1390Алгебра 8 клас Бевз 2025 · Повторення · стор. 264

1390. Розв’яжи рівняння. а) (x² − x)² − 11(x² − x) + 18 = 0; б) (√x − 2)² − 6(√x − 2) + 8 = 0.

а) ${\displaystyle {(x^2-x)}^2-11(x^2-x)+18=}$${\displaystyle 0}$;

заміна ${\displaystyle x^2-x=}$${\displaystyle t}$:

${\displaystyle t^2-11t+18=}$${\displaystyle 0}$;

${\displaystyle a=}$${\displaystyle 1,b=}$${\displaystyle -11,c=}$${\displaystyle 18}$;

${\displaystyle D=}$${\displaystyle b^2-4ac=}$${\displaystyle {(-11)}^2-4\cdot 1\cdot 18=}$${\displaystyle 121-72=}$${\displaystyle 49}$;

${\displaystyle D>0}$, два корені:

${\displaystyle t_{1,2}=}$${\displaystyle \frac{-b\pm \sqrt{D}}{2a}=}$${\displaystyle \frac{11\pm \sqrt{49}}{2}=}$${\displaystyle \frac{11\pm 7}{2}}$;

${\displaystyle t_1=}$${\displaystyle \frac{11+7}{2}=}$${\displaystyle 9}$;

${\displaystyle t_2=}$${\displaystyle \frac{11-7}{2}=}$${\displaystyle 2}$;

${\displaystyle x^2-x=}$${\displaystyle 9}$:

${\displaystyle x^2-x-9=}$${\displaystyle 0}$;

${\displaystyle a=}$${\displaystyle 1,b=}$${\displaystyle -1,c=}$${\displaystyle -9}$;

${\displaystyle D=}$${\displaystyle b^2-4ac=}$${\displaystyle {(-1)}^2-4\cdot 1\cdot (-9)=}$${\displaystyle 1+36=}$${\displaystyle 37}$;

${\displaystyle D>0}$, два корені:

${\displaystyle x_{1,2}=}$${\displaystyle \frac{-b\pm \sqrt{D}}{2a}=}$${\displaystyle \frac{1\pm \sqrt{37}}{2}}$;

${\displaystyle x^2-x=}$${\displaystyle 2}$:

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

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

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

${\displaystyle D>0}$, два корені:

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

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

${\displaystyle x_2=}$${\displaystyle \frac{1-3}{2}=}$${\displaystyle -1}$;

б) ${\displaystyle {(\sqrt{x}-2)}^2-6(\sqrt{x}-2)+8=}$${\displaystyle 0}$;

заміна ${\displaystyle \sqrt{x}-2=}$${\displaystyle t}$:

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

${\displaystyle t_1=}$${\displaystyle \frac{6+2}{2}=}$${\displaystyle 4}$;

${\displaystyle t_2=}$${\displaystyle \frac{6-2}{2}=}$${\displaystyle 2}$;

${\displaystyle \sqrt{x}-2=}$${\displaystyle 4}$;

${\displaystyle \sqrt{x}=}$${\displaystyle 6}$;

${\displaystyle x=}$${\displaystyle 36}$;

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

${\displaystyle \sqrt{x}=}$${\displaystyle 4}$;

${\displaystyle x=}$${\displaystyle 16}$.