827 Вправа 827Алгебра 8 клас Бевз 2025 · § 16 · стор. 156

827. Розклади на множники вираз.

а) ${\displaystyle \sqrt{35}-\sqrt{7}=}$${\displaystyle \sqrt{7}\cdot \sqrt{5}-\sqrt{7}=}$${\displaystyle \sqrt{7}\left(\sqrt{5}-1\right)}$;

б) ${\displaystyle a+\sqrt{a}=}$${\displaystyle \sqrt{a}\cdot \sqrt{a}+\sqrt{a}=}$${\displaystyle \sqrt{a}\left(\sqrt{a}+1\right)}$;

в) ${\displaystyle a\sqrt{c}-c\sqrt{a}}$=

=${\displaystyle \sqrt{a}\sqrt{c}\cdot \sqrt{a}-\sqrt{a}\sqrt{c}\cdot \sqrt{c}=}$${\displaystyle \sqrt{ac}\left(\sqrt{a}-\sqrt{c}\right)}$;

г) ${\displaystyle a^2-5=}$${\displaystyle \left(a-\sqrt{5}\right)\left(a+\sqrt{5}\right)}$;

ґ) ${\displaystyle p-a=}$${\displaystyle \left(\sqrt{p}-\sqrt{a}\right)\left(\sqrt{p}+\sqrt{a}\right)}$;

д) ${\displaystyle 9x^2-y}$=

=${\displaystyle {\left(3x\right)}^2-{\left(\sqrt{y}\right)}^2=}$${\displaystyle \left(3x-\sqrt{y}\right)\left(3x+\sqrt{y}\right)}$;

е) ${\displaystyle a+2\sqrt{a}+1=}$${\displaystyle {\left(\sqrt{a}\right)}^2+2\sqrt{a}+1=}$${\displaystyle {\left(\sqrt{a}+1\right)}^2}$;

є) ${\displaystyle x^2-6x\sqrt{c}+9c}$=

=${\displaystyle x^2-2\cdot x\cdot 3\sqrt{c}+{\left(3\sqrt{c}\right)}^2=}$${\displaystyle {\left(x-3\sqrt{c}\right)}^2}$.