629° Вправа 629Алгебра 8 клас Тарасенкова 2025 · § 15 · стор. 188

629. Розкрийте дужки та спростіть вираз:

1) ${\displaystyle \sqrt{p}(\sqrt{p}-15)=}$${\displaystyle {\left(\sqrt{p}\right)}^2-15\sqrt{p}=}$${\displaystyle p-15\sqrt{p}}$;

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

3) ${\displaystyle \sqrt{3}(\sqrt{a}+\sqrt{c})=}$${\displaystyle \sqrt{3a}+\sqrt{3c}}$;

4) ${\displaystyle (\sqrt{n}+7)(\sqrt{n}-2)=}$${\displaystyle {\left(\sqrt{n}\right)}^2-2\sqrt{n}+7\sqrt{n}-14=}$${\displaystyle n+5\sqrt{n}-14}$;

5) ${\displaystyle (2\sqrt{x}+\sqrt{y})(\sqrt{x}-2\sqrt{y})=}$${\displaystyle 2{\left(\sqrt{x}\right)}^2-4\sqrt{xy}+\sqrt{xy}-2{\left(\sqrt{y}\right)}^2=}$${\displaystyle 2x-3\sqrt{xy}-2y}$;

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

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

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

9) ${\displaystyle (\sqrt{5m}+\sqrt{n})(\sqrt{5m}-\sqrt{n})=}$${\displaystyle {\left(\sqrt{5m}\right)}^2-{\left(\sqrt{n}\right)}^2=}$${\displaystyle 5m-n}$;

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

11) ${\displaystyle {(\sqrt{a}-4)}^2=}$${\displaystyle {\left(\sqrt{a}\right)}^2-2\cdot 4\sqrt{a}+16=}$${\displaystyle a-8\sqrt{a}+16}$;

12) ${\displaystyle {(\sqrt{m}+\sqrt{3})}^2=}$${\displaystyle {\left(\sqrt{m}\right)}^2+2\sqrt{3m}+{\left(\sqrt{3}\right)}^2=}$${\displaystyle m+2\sqrt{3m}+3}$;

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

14) ${\displaystyle {(\sqrt{x}+\sqrt{z})}^2=}$${\displaystyle {\left(\sqrt{x}\right)}^2+2\sqrt{xz}+{\left(\sqrt{z}\right)}^2=}$${\displaystyle x+2\sqrt{xz}+z}$.