675* Вправа 675Алгебра 8 клас Тарасенкова 2025 · § 15 · стор. 196

675. Спростіть вираз:

1) ${\displaystyle x>2}$: ${\displaystyle \sqrt{{(x-1)}^2}=}$${\displaystyle |x-1|=}$${\displaystyle x-1}$;

2) ${\displaystyle x>20}$: ${\displaystyle \sqrt{{(x+12)}^2}=}$${\displaystyle |x+12|=}$${\displaystyle x+12}$;

3) ${\displaystyle x<0}$: ${\displaystyle \sqrt{{(x-8)}^2}=}$${\displaystyle |x-8|=}$${\displaystyle -(x-8)=}$${\displaystyle 8-x}$;

4) ${\displaystyle \sqrt{{(x^2+8)}^2}=}$${\displaystyle |x^2+8|=}$${\displaystyle x^2+8}$;

5) ${\displaystyle x>9}$: ${\displaystyle \sqrt{x^2-6x+9}=}$${\displaystyle \sqrt{{(x-3)}^2}=}$${\displaystyle |x-3|=}$${\displaystyle x-3}$;

6) ${\displaystyle x>5}$: ${\displaystyle \sqrt{x^2-2x+1}=}$${\displaystyle \sqrt{{(x-1)}^2}=}$${\displaystyle |x-1|=}$${\displaystyle x-1}$;

7) ${\displaystyle x<0}$: ${\displaystyle \sqrt{x^2-8x+16}=}$${\displaystyle \sqrt{{(x-4)}^2}=}$${\displaystyle |x-4|=}$${\displaystyle 4-x}$;

8) ${\displaystyle x<-4}$: ${\displaystyle \sqrt{x^2+4x+4}=}$${\displaystyle \sqrt{{(x+2)}^2}=}$${\displaystyle |x+2|=}$${\displaystyle -x-2}$.