559 Вправа 559Алгебра 8 клас Тарасенкова 2025 · § 13 · стор. 162

559. Спростіть вираз:

1) ${\displaystyle \sqrt{10}(\sqrt{28,9}-\sqrt{5,29})=}$${\displaystyle \sqrt{10}\cdot \sqrt{28,9}-\sqrt{10}\cdot \sqrt{5,29}=}$${\displaystyle \sqrt{289}-2,3\sqrt{10}=}$${\displaystyle 17-2,3\sqrt{10}}$;

2) ${\displaystyle \sqrt{72}=}$${\displaystyle \sqrt{36\cdot 2}=}$${\displaystyle 6\sqrt{2}}$; ${\displaystyle \sqrt{32}=}$${\displaystyle \sqrt{16\cdot 2}=}$${\displaystyle 4\sqrt{2}}$; ${\displaystyle \sqrt{8}=}$${\displaystyle 2\sqrt{2}}$;

${\displaystyle \frac{1}{\sqrt{8}}(\sqrt{72}+\sqrt{32})=}$${\displaystyle \frac{1}{2\sqrt{2}}(6\sqrt{2}+4\sqrt{2})=}$${\displaystyle \frac{10\sqrt{2}}{2\sqrt{2}}=}$${\displaystyle 5}$;

3) ${\displaystyle (\sqrt{1,3}+\sqrt{0,5})(\sqrt{1,3}-\sqrt{0,5})=}$${\displaystyle {\left(\sqrt{1,3}\right)}^2-{\left(\sqrt{0,5}\right)}^2=}$${\displaystyle 1,3-0,5=}$${\displaystyle 0,8}$;

4) ${\displaystyle \sqrt{{65}^2-{16}^2}=}$${\displaystyle \sqrt{(65-16)(65+16)}=}$${\displaystyle \sqrt{49\cdot 81}=}$${\displaystyle \sqrt{49}\cdot \sqrt{81}=}$${\displaystyle 7\cdot 9=}$${\displaystyle 63}$;

5) ${\displaystyle \sqrt{{37}^2-{12}^2}=}$${\displaystyle \sqrt{(37-12)(37+12)}=}$${\displaystyle \sqrt{25\cdot 49}=}$${\displaystyle \sqrt{25}\cdot \sqrt{49}=}$${\displaystyle 5\cdot 7=}$${\displaystyle 35}$;

6) ${\displaystyle {(\sqrt{10}+1)}^2+{(\sqrt{10}-1)}^2=}$${\displaystyle {\left(\sqrt{10}\right)}^2+2\sqrt{10}+1+{\left(\sqrt{10}\right)}^2-2\sqrt{10}+1=}$${\displaystyle 10+2\sqrt{10}+1+10-2\sqrt{10}+1=}$${\displaystyle 22}$;

7) ${\displaystyle {(\sqrt{6}+2)}^2-{(\sqrt{6}-2)}^2=}$${\displaystyle (6+4\sqrt{6}+4)-(6-4\sqrt{6}+4)=}$${\displaystyle 6+4\sqrt{6}+4-6+4\sqrt{6}-4=}$${\displaystyle 8\sqrt{6}}$;

8) ${\displaystyle {(\sqrt{2}+\sqrt{3})}^2+{(\sqrt{2}-\sqrt{3})}^2=}$${\displaystyle {\left(\sqrt{2}\right)}^2+2\sqrt{2}\sqrt{3}+{\left(\sqrt{3}\right)}^2+{\left(\sqrt{2}\right)}^2-2\sqrt{2}\sqrt{3}+{\left(\sqrt{3}\right)}^2=}$${\displaystyle 2+2\sqrt{6}+3+2-2\sqrt{6}+3=}$${\displaystyle 10}$.