1. \( \sqrt{\frac{16}{25}} \)
\( \sqrt{\frac{16}{25}} = \frac{\sqrt{16}}{\sqrt{25}} = \frac{4}{5} \)
2. \( \sqrt{9 \cdot 16} \)
\( \sqrt{9 \cdot 16} = \sqrt{9} \cdot \sqrt{16} = 3 \cdot 4 = 12 \)
3. \( \sqrt{\frac{9}{16}} \)
\( \sqrt{\frac{9}{16}} = \frac{\sqrt{9}}{\sqrt{16}} = \frac{3}{4} \)
4. \( \sqrt{\frac{49}{64}} \)
\( \sqrt{\frac{49}{64}} = \frac{\sqrt{49}}{\sqrt{64}} = \frac{7}{8} \)
5. \( \sqrt{\frac{36}{49}} \)
\( \sqrt{\frac{36}{49}} = \frac{\sqrt{36}}{\sqrt{49}} = \frac{6}{7} \)
6. \( \sqrt{49 \cdot a^2} \)
\( \sqrt{49 \cdot a^2} = \sqrt{49} \cdot \sqrt{a} = 7 \cdot |a| = 7|a| \)
magyarázat: a \( \sqrt{a^2} = |a| \), mert pl. \( \sqrt{4^2} = 4 \), és \( \sqrt{(-4)^2} = 4 \)
7. \( \sqrt{\frac{25}{81}} \)
\( \sqrt{\frac{25}{81}} = \frac{\sqrt{25}}{\sqrt{81}} = \frac{5}{9} \)
8. \( \sqrt{\frac{64}{100}} \)
\( \sqrt{\frac{64}{100}} = \frac{\sqrt{64}}{\sqrt{100}} = \frac{8}{10} = \frac{4}{5} \)
9. \( \sqrt{\frac{144}{225}} \)
\( \sqrt{\frac{144}{225}} = \frac{\sqrt{144}}{\sqrt{225}} = \frac{12}{15} = \frac{4}{5} \)
10. \( \sqrt{\frac{289}{256}} \)
\( \sqrt{\frac{289}{256}} = \frac{\sqrt{289}}{\sqrt{256}} = \frac{17}{16} \)
11. \( \sqrt{\frac{81}{169}} \)
\( \sqrt{\frac{81}{169}} = \frac{\sqrt{81}}{\sqrt{169}} = \frac{9}{13} \)
12. \( \sqrt{\frac{25}{49}} \)
\( \sqrt{\frac{25}{49}} = \frac{\sqrt{25}}{\sqrt{49}} = \frac{5}{7} \)
13. \( \sqrt{121 \cdot 256} \)
\( \sqrt{121 \cdot 256} = \sqrt{121} \cdot \sqrt{256} = 11 \cdot 16 = 176 \)
14. \( \sqrt{\frac{169}{361}} \)
\( \sqrt{\frac{169}{361}} = \frac{\sqrt{169}}{\sqrt{361}} = \frac{13}{19} \)
15. \( \sqrt{\frac{324}{289}} \)
\( \sqrt{\frac{324}{289}} = \frac{\sqrt{324}}{\sqrt{289}} = \frac{18}{17} \)
1. \( \sqrt{8} + \sqrt{18} \)
\( \sqrt{8} = 2\sqrt{2}, \sqrt{18} = 3\sqrt{2} \)
Tehát: \( 2\sqrt{2} + 3\sqrt{2} = 5\sqrt{2} \)
2. \( \sqrt{12} - \sqrt{27} \)
\( \sqrt{12} = 2\sqrt{3}, \sqrt{27} = 3\sqrt{3} \)
Tehát: \( 2\sqrt{3} - 3\sqrt{3} = -\sqrt{3} \)
3. \( 2\sqrt{5} + 3\sqrt{5} \)
Tehát: \( 2\sqrt{5} + 3\sqrt{5} = 5\sqrt{5} \)
4. \( \sqrt{50} - \sqrt{18} \)
\( \sqrt{50} = 5\sqrt{2}, \sqrt{18} = 3\sqrt{2} \)
Tehát: \( 5\sqrt{2} - 3\sqrt{2} = 2\sqrt{2} \)
5. \( \sqrt{72} + \sqrt{50} \)
\( \sqrt{72} = 6\sqrt{2}, \sqrt{50} = 5\sqrt{2} \)
Tehát: \( 6\sqrt{2} + 5\sqrt{2} = 11\sqrt{2} \)
6. \( 3\sqrt{7} - 2\sqrt{7} \)
Tehát: \( 3\sqrt{7} - 2\sqrt{7} = \sqrt{7} \)
7. \( \sqrt{32} + \sqrt{18} \)
\( \sqrt{32} = 4\sqrt{2}, \sqrt{18} = 3\sqrt{2} \)
Tehát: \( 4\sqrt{2} + 3\sqrt{2} = 7\sqrt{2} \)
8. \( \sqrt{75} - \sqrt{12} \)
\( \sqrt{75} = 5\sqrt{3}, \sqrt{12} = 2\sqrt{3} \)
Tehát: \( 5\sqrt{3} - 2\sqrt{3} = 3\sqrt{3} \)
9. \( 7\sqrt{5} + 2\sqrt{5} \)
Tehát: \( 7\sqrt{5} + 2\sqrt{5} = 9\sqrt{5} \)
10. \( \sqrt{45} - \sqrt{20} \)
\( \sqrt{45} = 3\sqrt{5}, \sqrt{20} = 2\sqrt{5} \)
Tehát: \( 3\sqrt{5} - 2\sqrt{5} = \sqrt{5} \)
1. \( \sqrt{50} \)
\( \sqrt{50} = \sqrt{25 \cdot 2} = 5\sqrt{2} \)
2. \( \sqrt{72} \)
\( \sqrt{72} = \sqrt{36 \cdot 2} = 6\sqrt{2} \)
3. \( \sqrt{125} \)
\( \sqrt{125} = \sqrt{25 \cdot 5} = 5\sqrt{5} \)
4. \( \sqrt{200} \)
\( \sqrt{200} = \sqrt{100 \cdot 2} = 10\sqrt{2} \)
5. \( \sqrt{45} \)
\( \sqrt{45} = \sqrt{9 \cdot 5} = 3\sqrt{5} \)
6. \( \sqrt{32} \)
\( \sqrt{32} = \sqrt{16 \cdot 2} = 4\sqrt{2} \)
7. \( \sqrt{180} \)
\( \sqrt{180} = \sqrt{36 \cdot 5} = 6\sqrt{5} \)
8. \( \sqrt{50} \)
\( \sqrt{50} = \sqrt{25 \cdot 2} = 5\sqrt{2} \)
9. \( \sqrt{18} \)
\( \sqrt{18} = \sqrt{9 \cdot 2} = 3\sqrt{2} \)
10. \( \sqrt{300} \)
\( \sqrt{300} = \sqrt{100 \cdot 3} = 10\sqrt{3} \)
1. \( 2\sqrt{5} \)
\( 2\sqrt{5} = \sqrt{4 \cdot 5} = \sqrt{20} \)
2. \( 3\sqrt{7} \)
\( 3\sqrt{7} = \sqrt{9 \cdot 7} = \sqrt{63} \)
3. \( 4\sqrt{2} \)
\( 4\sqrt{2} = \sqrt{16 \cdot 2} = \sqrt{32} \)
4. \( 5\sqrt{3} \)
\( 5\sqrt{3} = \sqrt{25 \cdot 3} = \sqrt{75} \)
5. \( 6\sqrt{6} \)
\( 6\sqrt{6} = \sqrt{36 \cdot 6} = \sqrt{216} \)
6. \( 7\sqrt{5} \)
\( 7\sqrt{5} = \sqrt{49 \cdot 5} = \sqrt{245} \)
7. \( 8\sqrt{2} \)
\( 8\sqrt{2} = \sqrt{64 \cdot 2} = \sqrt{128} \)
8. \( 9\sqrt{3} \)
\( 9\sqrt{3} = \sqrt{81 \cdot 3} = \sqrt{243} \)
9. \( 10\sqrt{7} \)
\( 10\sqrt{7} = \sqrt{100 \cdot 7} = \sqrt{700} \)
10. \( 11\sqrt{2} \)
\( 11\sqrt{2} = \sqrt{121 \cdot 2} = \sqrt{242} \)
1. \( 6\sqrt{3} \) vagy \( 5\sqrt{5} \)?
\( 6\sqrt{3} = \sqrt{36\cdot3} = \sqrt{108} \)
\( 5\sqrt{5} = \sqrt{25\cdot5} = \sqrt{125} \), tehát az \( 5\sqrt{5} \) a nagyobb.
2. \( 5\sqrt{2} \) vagy \( \sqrt{45} \)?
\( 5\sqrt{2} = \sqrt{25\cdot2} = \sqrt{50} \) tehát az \( 5\sqrt{2} \) a nagyobb.
3. \( 2\sqrt{7} \) vagy \( 3\sqrt{5} \)?
\( 2\sqrt{7} = \sqrt{4\cdot7} = \sqrt{28} \)
\( 3\sqrt{5} = \sqrt{9\cdot5} = \sqrt{45} \), tehát a \( 3\sqrt{5} \) a nagyobb.
4. \( \sqrt{72} \) vagy \( 6\sqrt{2} \)?
\( \sqrt{72} = 6\sqrt{2} \), tehát egyenlőek.
5. \( 4\sqrt{3} \) vagy \( 5\sqrt{2} \)?
\( 4\sqrt{3} = \sqrt{16\cdot3} = \sqrt{48} \)
\( 5\sqrt{2} = \sqrt{25\cdot2} = \sqrt{50} \), tehát az \( 5\sqrt{2} \) a nagyobb.
6. \( 3\sqrt{7} \) vagy \( 4\sqrt{5} \)?
\( 3\sqrt{7} = \sqrt{9\cdot7} = \sqrt{63} \)
\( 4\sqrt{5} = \sqrt{16\cdot5} = \sqrt{80} \), tehát a \( 4\sqrt{5} \) a nagyobb.
7. \( 2\sqrt{10} \) vagy \( 3\sqrt{6} \)?
\( 2\sqrt{10} = \sqrt{4\cdot10} = \sqrt{40} \)
\( 3\sqrt{6} = \sqrt{9\cdot6} = \sqrt{54} \), tehát a \( 3\sqrt{6} \) a nagyobb.
8. \( 6\sqrt{3} \) vagy \( 4\sqrt{5} \)?
\( 6\sqrt{3} = \sqrt{36\cdot3} = \sqrt{108} \)
\( 4\sqrt{5} = \sqrt{16\cdot5} = \sqrt{80} \), tehát a \( 6\sqrt{3} \) a nagyobb.
9. \( 3\sqrt{11} \) vagy \( 7\sqrt{2} \)?
\( 3\sqrt{11} = \sqrt{9\cdot11} = \sqrt{99} \)
\( 7\sqrt{2} = \sqrt{49\cdot2} = \sqrt{98} \), tehát a \( 3\sqrt{11} \) a nagyobb.
10. \( 8\sqrt{2} \) vagy \( 7\sqrt{3} \)?
\( 8\sqrt{2} = \sqrt{64\cdot2} = \sqrt{128} \)
\( 7\sqrt{3} = \sqrt{49\cdot3} = \sqrt{147} \), tehát a \( 7\sqrt{3} \) a nagyobb.
11. \( \frac{5\sqrt{3}}{\sqrt{2}} \) vagy \( \frac{7\sqrt{2}}{\sqrt{3}} \)?
\( \frac{5\sqrt{3}}{\sqrt{2}} = \sqrt{\frac{25\cdot3}{2}} = \sqrt{\frac{75}{2}} = 37,5 \)
\( \frac{7\sqrt{2}}{\sqrt{3}} = \sqrt{\frac{49\cdot2}{3}} = \sqrt{\frac{98}{3}} \approx 33 \), tehát az \( \frac{5\sqrt{3}}{\sqrt{2}} \) a nagypbb.
12. \( \frac{\sqrt{33}}{2} \) vagy \( \frac{\sqrt{52}}{3} \)?
\( \frac{\sqrt{33}}{2} = \sqrt{\frac{33}{4}} > \sqrt{8} \)
\( \frac{\sqrt{52}}{3} = \sqrt{\frac{52}{9}} < \sqrt{6} \), tehát a \( \frac{\sqrt{33}}{2} \) a nagyobb.
1. \( (\sqrt{5} + 2)^2 \)
Használjuk a nevezetes azonosságot: \( (a+b)^2 = a^2 + 2ab + b^2 \)
\( (\sqrt{5} + 2)^2 = (\sqrt{5})^2 + 2 \cdot \sqrt{5} \cdot 2 + 2^2 = 5 + 4\sqrt{5} + 4 = 9 + 4\sqrt{5} \)
2. \( (\sqrt{3} - 4)^2 \)
\( (\sqrt{3} - 4)^2 = (\sqrt{3})^2 - 2 \cdot \sqrt{3} \cdot 4 + 4^2 = 3 - 8\sqrt{3} + 16 = 19 - 8\sqrt{3} \)
3. \( (2\sqrt{2} + 3)^2 \)
\( (2\sqrt{2} + 3)^2 = (2\sqrt{2})^2 + 2 \cdot 2\sqrt{2} \cdot 3 + 3^2 = 8 + 12\sqrt{2} + 9 = 17 + 12\sqrt{2} \)
4. \( (3 - \sqrt{7})^2 \)
\( (3 - \sqrt{7})^2 = 3^2 - 2 \cdot 3 \cdot \sqrt{7} + (\sqrt{7})^2 = 9 - 6\sqrt{7} + 7 = 16 - 6\sqrt{7} \)
5. \( (5 + \sqrt{3})^2 \)
\( (5 + \sqrt{3})^2 = 5^2 + 2 \cdot 5 \cdot \sqrt{3} + (\sqrt{3})^2 = 25 + 10\sqrt{3} + 3 = 28 + 10\sqrt{3} \)
6. \( (4\sqrt{3} - 1)^2 \)
\( (4\sqrt{3} - 1)^2 = (4\sqrt{3})^2 - 2 \cdot 4\sqrt{3} \cdot 1 + 1^2 = 48 - 8\sqrt{3} + 1 = 49 - 8\sqrt{3} \)
7. \( (\sqrt{6} + 2)^2 \)
\( (\sqrt{6} + 2)^2 = (\sqrt{6})^2 + 2 \cdot \sqrt{6} \cdot 2 + 2^2 = 6 + 4\sqrt{6} + 4 = 10 + 4\sqrt{6} \)
8. \( (3\sqrt{5} - 2)^2 \)
\( (3\sqrt{5} - 2)^2 = (3\sqrt{5})^2 - 2 \cdot 3\sqrt{5} \cdot 2 + 2^2 = 45 - 12\sqrt{5} + 4 = 49 - 12\sqrt{5} \)
9. \( (2\sqrt{7} + 5)^2 \)
\( (2\sqrt{7} + 5)^2 = (2\sqrt{7})^2 + 2 \cdot 2\sqrt{7} \cdot 5 + 5^2 = 28 + 20\sqrt{7} + 25 = 53 + 20\sqrt{7} \)
10. \( (5 - \sqrt{2})^2 \)
\( (5 - \sqrt{2})^2 = 5^2 - 2 \cdot 5 \cdot \sqrt{2} + (\sqrt{2})^2 = 25 - 10\sqrt{2} + 2 = 27 - 10\sqrt{2} \)
11. \( (\sqrt{3} - 2) \cdot (\sqrt{3} + 2) \)
\( (\sqrt{3})^2 - 2^2 = 3 - 4 = -1 \)
12. \( (7 + \sqrt{2})^2 \cdot (7 - \sqrt{2}) \)
\( 7^2 - (\sqrt{2})^2 = 49 - 2 = 47 \)
13. \( (4 + \sqrt{5})\cdot(4 - \sqrt{5}) \)
\( 4^2 - (\sqrt{5})^2 = 16 - 5 = 11 \)
14. \( (\sqrt{8} + 3)\cdot(\sqrt{8} - 3) \)
\( (\sqrt{8})^2 - 3^2 = 8 - 9 = -1 \)
15. \( (\sqrt{a} - 7)\cdot(\sqrt{a} + 7) \)
\( (\sqrt{a})^2 - 7^2 = a - 49 \)
16. \( (\sqrt{x} + 3)\cdot(\sqrt{x} - 3) \)
\( (\sqrt{x})^2 - 3^2 = x - 9 \)
17. Alakítsd szorzattá!
\( x^2 - 5 \)
\( x^2 - 5 = (\sqrt{x} - 5)\cdot(\sqrt{x} + 5) \)
18. Alakítsd szorzattá!
\( 144 - a^2 \)
\( 144 -a^2 = (12 - a)\cdot(12 + a) \)
19. Alakítsd szorzattá!
\( 169 - x \)
\( 169 - x = (13 + \sqrt{x})\cdot(13 - \sqrt{x}) \)
20. Alakítsd szorzattá!
\( x - 4 \)
\( x - 4 = (\sqrt{x} - 2)\cdot(\sqrt{x} + 2) \)
21. Alakítsd szorzattá!
\( 4a - b \)
\( 4a - b = (2\sqrt{a} - b)\cdot(2\sqrt{a} + b) \)