Matematică, întrebare adresată de gau59, 8 ani în urmă

Am si eu nevoie de ajutor la aceste 3 exercitii! Multumesc!

Anexe:

Răspunsuri la întrebare

Răspuns de Rayzen
2

21)\\ \\ E = \sqrt{2-\sqrt{3}+\sqrt{5-\sqrt{13+\sqrt{48}}}}\in \mathbb{Q}\,?\\ \\ \sqrt{13+\sqrt{48}} = \sqrt{13+4\sqrt{3}} = \sqrt{(2\sqrt{3}+1)^2} = |2\sqrt 3+1| = 1+2\sqrt{3}\\ \sqrt{5-1-2\sqrt{3}} = \sqrt{4-2\sqrt 3} = \sqrt{(\sqrt{3}-1)^2} = |\sqrt{3}-1| = \sqrt{3}-1\\ \\\\ \Rightarrow E = \sqrt{2-\sqrt{3}+\sqrt{3}-1} = \sqrt{1} = 1\in \mathbb{Q}

\\E = \sqrt{26-6\sqrt{13+4\sqrt{8-2\sqrt{6+2\sqrt{5}}}}} \in \mathbb{R}\,\backslash\,\mathbb{Q}\,?\\ \\ \sqrt{6+2\sqrt{5}} = \sqrt{(\sqrt{5}+1)^2} = \sqrt{5}+1\\ \sqrt{8-2(\sqrt{5}+1)} = \sqrt{6-2\sqrt{5}} = \sqrt{(\sqrt{5}-1)^2} = |\sqrt{5}-1| = \sqrt{5}-1 \\ \sqrt{13+4(\sqrt{5}-1)} = \sqrt{9-4\sqrt{5}} = |2-\sqrt{5}| = \sqrt{5}-2 \\ \\\Rightarrow E = \sqrt{26-6(\sqrt{5}-2)} = \sqrt{14-6\sqrt{5}} =\sqrt{(3-\sqrt{5})^2} = |3-\sqrt{5}|\\ \\ \Rightarrow E = 3-\sqrt{5}\\\\ \Rightarrow E \in \mathbb{R}\,\backslash\,\mathbb{Q}

\\22)\\ \\ x = \sqrt{441+2+4+6+...+880}\\ x = \sqrt{441+2(1+2+3+...+440)}\\ x = \sqrt{441+2\cdot \dfrac{440\cdot 441}{2}}\\ x = \sqrt{441+440\cdot 441}\\ x = \sqrt{441\cdot(440+1)}\\ x = \sqrt{441^2}\\ x = 441\\ x = 21^2\quad \Rightarrow \quad c)\,\,\sqrt{x}\in \mathbb{Z}

\\23)\\ \\ x = \sqrt{243^2-(240^2+3\cdot 240)}\\ x = \sqrt{243^2 - \big[240\cdot(240+3)\big]}\\x =\sqrt{243^2 - 240\cdot 243}\\ x = \sqrt{243\cdot (243 - 240)}\\ x = \sqrt{243\cdot 3}\\ x = \sqrt{3\cdot 81\cdot 3}\\x = \sqrt{3^2\cdot 9^2}\\ x =3\cdot 9 \\ x = 27\\ x = (3\sqrt 3)^2\quad \Rightarrow\quad c)\,\,\sqrt{x}\in \mathbb{R}\,\backslash\,\mathbb{Q}

Alte întrebări interesante