571 Вправа 571Алгебра 8 клас Тарасенкова 2025 · § 13 · стор. 164

571. Спростіть вираз:

1) ${\displaystyle 2\sqrt{2}+\sqrt{8}=}$${\displaystyle 2\sqrt{2}+\sqrt{4\cdot 2}=}$${\displaystyle 2\sqrt{2}+2\sqrt{2}=}$${\displaystyle (2+2)\sqrt{2}=}$${\displaystyle 4\sqrt{2}}$;

2) ${\displaystyle 5\sqrt{12}-\sqrt{27}=}$${\displaystyle 5\sqrt{4\cdot 3}-\sqrt{9\cdot 3}=}$${\displaystyle 5\cdot 2\sqrt{3}-3\sqrt{3}=}$${\displaystyle 10\sqrt{3}-3\sqrt{3}=}$${\displaystyle (10-3)\sqrt{3}=}$${\displaystyle 7\sqrt{3}}$;

3) ${\displaystyle \sqrt{5}+\sqrt{20}+\sqrt{45}=}$${\displaystyle \sqrt{5}+\sqrt{4\cdot 5}+\sqrt{9\cdot 5}=}$${\displaystyle \sqrt{5}+2\sqrt{5}+3\sqrt{5}=}$${\displaystyle (1+2+3)\sqrt{5}=}$${\displaystyle 6\sqrt{5}}$;

4) ${\displaystyle 3\sqrt{10}-\sqrt{40}+\sqrt{90}=}$${\displaystyle 3\sqrt{10}-\sqrt{4\cdot 10}+\sqrt{9\cdot 10}=}$${\displaystyle 3\sqrt{10}-2\sqrt{10}+3\sqrt{10}=}$${\displaystyle (3-2+3)\sqrt{10}=}$${\displaystyle 4\sqrt{10}}$;

5) ${\displaystyle \sqrt{12}+\sqrt{27}+\sqrt{48}=}$${\displaystyle \sqrt{4\cdot 3}+\sqrt{9\cdot 3}+\sqrt{16\cdot 3}=}$${\displaystyle 2\sqrt{3}+3\sqrt{3}+4\sqrt{3}=}$${\displaystyle (2+3+4)\sqrt{3}=}$${\displaystyle 9\sqrt{3}}$;

6) ${\displaystyle \sqrt{9,8}-0,2\sqrt{45}+\sqrt{2,45}=}$${\displaystyle \sqrt{1,96\cdot 5}-0,2\sqrt{9\cdot 5}+\sqrt{0,49\cdot 5}=}$${\displaystyle 1,4\sqrt{5}-0,2\cdot 3\sqrt{5}+0,7\sqrt{5}=}$${\displaystyle 1,4\sqrt{5}-0,6\sqrt{5}+0,7\sqrt{5}=}$${\displaystyle (1,4-0,6+0,7)\sqrt{5}=}$${\displaystyle 1,5\sqrt{5}}$.