863* Вправа 863Алгебра 8 клас Тарасенкова 2025 · § 21 · стор. 256

863. Розв’яжіть рівняння:

1) ${\displaystyle x^4-(n^2+9)x^2+9n^2=}$${\displaystyle 0}$; нехай ${\displaystyle x^2=}$${\displaystyle t}$, ${\displaystyle t\ge 0}$:

${\displaystyle t^2-(n^2+9)t+9n^2=}$${\displaystyle 0}$;

${\displaystyle D=}$${\displaystyle {(-(n^2+9))}^2-4\cdot 1\cdot 9n^2=}$${\displaystyle n^4+18n^2+81-36n^2}$;

${\displaystyle D=}$${\displaystyle n^4-18n^2+81=}$${\displaystyle {(n^2-9)}^2}$;

${\displaystyle t_1=}$${\displaystyle \frac{n^2+9-(n^2-9)}{2}=}$${\displaystyle \frac{18}{2}=}$${\displaystyle 9}$;

${\displaystyle t_2=}$${\displaystyle \frac{n^2+9+n^2-9}{2}=}$${\displaystyle \frac{2n^2}{2}=}$${\displaystyle n^2}$;

${\displaystyle x^2=}$${\displaystyle 9}$ або ${\displaystyle x^2=}$${\displaystyle n^2}$;

${\displaystyle x_1=}$${\displaystyle 3}$; ${\displaystyle x_2=}$${\displaystyle -3}$; ${\displaystyle x_3=}$${\displaystyle n}$; ${\displaystyle x_4=}$${\displaystyle -n}$;

2) ${\displaystyle x+\frac{1}{n}=}$${\displaystyle n+\frac{1}{x}}$; ОДЗ: ${\displaystyle n\ne 0}$, ${\displaystyle x\ne 0}$;

${\displaystyle x-\frac{1}{x}+\frac{1}{n}-n=}$${\displaystyle 0}$;

${\displaystyle \frac{x^2-1}{x}+\frac{1-n^2}{n}=}$${\displaystyle 0}$;

${\displaystyle \frac{n(x^2-1)+x(1-n^2)}{nx}=}$${\displaystyle 0}$;

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

${\displaystyle (n{x}^2-x{n}^2)+(x-n)=}$${\displaystyle 0}$;

${\displaystyle nx(x-n)+1\cdot (x-n)=}$${\displaystyle 0}$;

${\displaystyle (x-n)(nx+1)=}$${\displaystyle 0}$;

${\displaystyle x-n=}$${\displaystyle 0}$ або ${\displaystyle nx+1=}$${\displaystyle 0}$;

${\displaystyle x_1=}$${\displaystyle n}$; ${\displaystyle x_2=}$${\displaystyle -\frac{1}{n}}$.