829* Вправа 829Алгебра 8 клас Тарасенкова 2025 · § 20 · стор. 244

829. Розкладіть на лінійні множники квадратний тричлен:

1)

${\displaystyle D=}$${\displaystyle {(n-2)}^2-4\cdot n\cdot (-2)=}$${\displaystyle n^2-4n+4+8n}$;

${\displaystyle D=}$${\displaystyle n^2+4n+4=}$${\displaystyle {(n+2)}^2}$;

${\displaystyle x_1=}$${\displaystyle \frac{-(n-2)-\sqrt{{(n+2)}^2}}{2n}=}$${\displaystyle \frac{-n+2-(n+2)}{2n}=}$${\displaystyle \frac{-2n}{2n}=}$${\displaystyle -1}$;

${\displaystyle x_2=}$${\displaystyle \frac{-(n-2)+\sqrt{{(n+2)}^2}}{2n}=}$${\displaystyle \frac{-n+2+n+2}{2n}=}$${\displaystyle \frac{4}{2n}=}$${\displaystyle \frac{2}{n}}$;

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

2)

${\displaystyle D=}$${\displaystyle {(-m)}^2-4\cdot 1\cdot (-mn-n^2)=}$${\displaystyle m^2+4mn+4n^2=}$${\displaystyle {(m+2n)}^2}$;

${\displaystyle x_1=}$${\displaystyle \frac{m-\sqrt{{(m+2n)}^2}}{2\cdot 1}=}$${\displaystyle \frac{m-(m+2n)}{2}=}$${\displaystyle \frac{-2n}{2}=}$${\displaystyle -n}$;

${\displaystyle x_2=}$${\displaystyle \frac{m+\sqrt{{(m+2n)}^2}}{2\cdot 1}=}$${\displaystyle \frac{m+m+2n}{2}=}$${\displaystyle \frac{2m+2n}{2}=}$${\displaystyle m+n}$;

${\displaystyle x^2-mx-mn-n^2=}$${\displaystyle (x-(m+n))(x+n)=}$${\displaystyle \left(x-m-n\right)\left(x+n\right)}$.