544° Вправа 544Алгебра 8 клас Тарасенкова 2025 · § 13 · стор. 160

544. Спростіть вираз:

1) ${\displaystyle \sqrt{3}(\sqrt{3}+1)=}$${\displaystyle {\left(\sqrt{3}\right)}^2+\sqrt{3}=}$${\displaystyle 3+\sqrt{3}}$;

2) ${\displaystyle \sqrt{5}(6+\sqrt{5})=}$${\displaystyle 6\sqrt{5}+{\left(\sqrt{5}\right)}^2=}$${\displaystyle 6\sqrt{5}+5}$;

3) ${\displaystyle (\sqrt{8}+1)(\sqrt{2}-2)=}$${\displaystyle \sqrt{16}-2\sqrt{8}+\sqrt{2}-2=}$${\displaystyle 4-4\sqrt{2}+\sqrt{2}-2=}$${\displaystyle 2-3\sqrt{2}}$;

4) ${\displaystyle (\sqrt{15}+4)(\sqrt{15}-4)=}$${\displaystyle {\left(\sqrt{15}\right)}^2-4^2=}$${\displaystyle 15-16=}$${\displaystyle -1}$;

5) ${\displaystyle (\sqrt{5}+1)(\sqrt{5}-1)=}$${\displaystyle {\left(\sqrt{5}\right)}^2-1^2=}$${\displaystyle 5-1=}$${\displaystyle 4}$;

6) ${\displaystyle (\sqrt{21}+\sqrt{11})(\sqrt{21}-\sqrt{11})=}$${\displaystyle {\left(\sqrt{21}\right)}^2-{\left(\sqrt{11}\right)}^2=}$${\displaystyle 21-11=}$${\displaystyle 10}$;

7) ${\displaystyle {(\sqrt{2}-1)}^2=}$${\displaystyle {\left(\sqrt{2}\right)}^2-2\sqrt{2}+1=}$${\displaystyle 2-2\sqrt{2}+1=}$${\displaystyle 3-2\sqrt{2}}$;

8) ${\displaystyle {(\sqrt{5}-2)}^2=}$${\displaystyle {\left(\sqrt{5}\right)}^2-4\sqrt{5}+4=}$${\displaystyle 5-4\sqrt{5}+4=}$${\displaystyle 9-4\sqrt{5}}$.