630° Вправа 630Алгебра 8 клас Тарасенкова 2025 · § 15 · стор. 189

630. Розкрийте дужки та спростіть вираз:

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

2) ${\displaystyle (\sqrt{a}+3)(\sqrt{a}-6)=}$${\displaystyle {\left(\sqrt{a}\right)}^2-6\sqrt{a}+3\sqrt{a}-18=}$${\displaystyle a-3\sqrt{a}-18}$;

3) ${\displaystyle (\sqrt{y}+8)(\sqrt{y}-8)=}$${\displaystyle {\left(\sqrt{y}\right)}^2-8^2=}$${\displaystyle y-64}$;

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

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

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

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

8) ${\displaystyle {(\sqrt{y}+\sqrt{10})}^2=}$${\displaystyle {\left(\sqrt{y}\right)}^2+2\sqrt{10y}+{\left(\sqrt{10}\right)}^2=}$${\displaystyle y+2\sqrt{10y}+10}$.