764 Вправа 764Алгебра 8 клас Бевз 2025 · § 15 · стор. 146

764. Спрости вираз.

а) ${\displaystyle \sqrt{16m^2}=}$${\displaystyle 4\left|m\right|}$;

якщо ${\displaystyle m<0}$ — ${\displaystyle 4\left|m\right|=}$${\displaystyle -4m}$;

б) ${\displaystyle \sqrt{{(-x)}^2}=}$${\displaystyle |-x|=}$${\displaystyle \left|x\right|}$;

якщо ${\displaystyle x>0}$ — ${\displaystyle x}$;

в) ${\displaystyle -0,2a\sqrt{64a^{14}}=}$${\displaystyle -0,2a\cdot 8\left|a^7\right|}$;

якщо ${\displaystyle a\le 0}$, то ${\displaystyle \left|a^7\right|=}$${\displaystyle -a^7}$;

${\displaystyle -0,2a\cdot 8\cdot (-a^7)=}$${\displaystyle 1,6a^8}$;

г) ${\displaystyle \sqrt{0,81a^6{b}^6}=}$${\displaystyle 0,9\left|a^3\right|\cdot \left|b^3\right|}$;

якщо ${\displaystyle a\ge 0}$, ${\displaystyle b\le 0}$, то ${\displaystyle \left|a^3\right|=}$${\displaystyle a^3}$;

${\displaystyle \left|b^3\right|=}$${\displaystyle -b^3}$;

${\displaystyle 0,9\cdot a^3\cdot (-b^3)=}$${\displaystyle -0,9a^3{b}^3}$;

ґ) ${\displaystyle \sqrt{{(a-b)}^2}=}$${\displaystyle |a-b|}$;

якщо ${\displaystyle a<b}$, то ${\displaystyle a-b<0}$ — ${\displaystyle |a-b|=}$${\displaystyle b-a}$;

д) ${\displaystyle \sqrt{{(x+5)}^2}=}$${\displaystyle |x+5|}$;

якщо ${\displaystyle x>-5}$, то ${\displaystyle x+5>0}$ — ${\displaystyle |x+5|=}$${\displaystyle x+5}$.