Zadania matematyczne

Zadanie 4

Wyłączanie całości przed znak pierwiastka:

\(\sqrt{8} + 3\sqrt{2} =\)

\(\sqrt{8} = \sqrt{4\cdot 2} = 2\sqrt{2}\)

\(3\sqrt{3} - 2\sqrt{27} =\)

\(\sqrt{27} = \sqrt{9\cdot 3} = 3\sqrt{3}\)

\(\sqrt{50} - \sqrt{8} =\)

\(\sqrt{50}= \sqrt{25\cdot 2}=5\sqrt{2}, \; \sqrt{8}=2\sqrt{2}\)

\(\sqrt{12} + 2\sqrt{27}=\)

\(\sqrt{12} = 2\sqrt{3}, \; \sqrt{27} = 3\sqrt{3}\)

\(-2\sqrt{18} + \sqrt{8}=\)

\(\sqrt{18}=3\sqrt{2}, \; \sqrt{8}=2\sqrt{2}\)

\(4\sqrt{20} - \sqrt{45}=\)

\(\sqrt{20}=2\sqrt{5}, \; \sqrt{45}=3\sqrt{5}\)

\(\sqrt{72} + 2\sqrt{8}=\)

\(\sqrt{72}=6\sqrt{2}, \; \sqrt{8}=2\sqrt{2}\)

\(5\sqrt{2} - \sqrt{18}=\)

\(\sqrt{18}=3\sqrt{2}\)

\(\sqrt{8}+3\sqrt{2} = 2\sqrt{2}+3\sqrt{2}=5\sqrt{2}\)

\(3\sqrt{3}-2\sqrt{27}=3\sqrt{3}-6\sqrt{3}=-3\sqrt{3}\)

\(\sqrt{50}-\sqrt{8}=5\sqrt{2}-2\sqrt{2}=3\sqrt{2}\)

\(\sqrt{12}+2\sqrt{27}=2\sqrt{3}+6\sqrt{3}=8\sqrt{3}\)

\(-2\sqrt{18}+\sqrt{8}=-6\sqrt{2}+2\sqrt{2}=-4\sqrt{2}\)

\(4\sqrt{20}-\sqrt{45}=8\sqrt{5}-3\sqrt{5}=5\sqrt{5}\)

\(\sqrt{72}+2\sqrt{8}=6\sqrt{2}+4\sqrt{2}=10\sqrt{2}\)

\(5\sqrt{2}-\sqrt{18}=5\sqrt{2}-3\sqrt{2}=2\sqrt{2}\)

Kartkówka – Grupa 1

Kartkówka – Grupa 2

Kartkówka – Grupa 3