Działania na pierwiastkach i ich upraszczanie
Wyobraź sobie, że pierwiastki to paczki w magazynie. Nie zawsze da się je dodać „wprost”, ale często można je przepakować tak, żeby były lżejsze i łatwiejsze w obliczeniach. To właśnie nazywamy upraszczaniem.
Najważniejsza własność (mnożenie i dzielenie)
Dla pierwiastków tego samego stopnia (i w przypadku kwadratowych: dla liczb nieujemnych) działa:
- \(\sqrt{a}\cdot\sqrt{b}=\sqrt{ab}\)
- \(\sqrt{a}:\sqrt{b}=\sqrt{\frac{a}{b}}\) (dla \(b\neq 0\))
Przykłady (mnożenie i dzielenie)
- \(\sqrt{2}\cdot\sqrt{8}=\sqrt{16}=4\)
- \(\sqrt{27}:\sqrt{3}=\sqrt{9}=3\)
- \(\sqrt[3]{2}\cdot\sqrt[3]{4}=\sqrt[3]{8}=2\)
Upraszczanie: „wyciąganie kwadratu” spod pierwiastka
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Jeśli pod pierwiastkiem widzisz iloczyn, w którym jest pełny kwadrat, możesz go wyciągnąć przed znak pierwiastka:
\[\sqrt{50}=\sqrt{25\cdot 2}=\sqrt{25}\cdot\sqrt{2}=5\sqrt{2}\]
To działa, bo \(25=5^2\).
Przykład 1
Uprość:
- \(\sqrt{98}\)
- \(2\sqrt{5}\cdot\sqrt{10}\)
- \(\dfrac{\sqrt{32}}{\sqrt{2}}\)
Rozwiązanie:
- \(\sqrt{98}=\sqrt{49\cdot 2}=7\sqrt{2}\)
- \(2\sqrt{5}\cdot\sqrt{10}=2\sqrt{50}=2\cdot 5\sqrt{2}=10\sqrt{2}\)
- \(\dfrac{\sqrt{32}}{\sqrt{2}}=\sqrt{\dfrac{32}{2}}=\sqrt{16}=4\)
Włączanie czynnika pod pierwiastek (ruch w drugą stronę)
Czasem wygodniej jest „schować” liczbę pod pierwiastkiem:
\[3\sqrt{7}=\sqrt{3^2\cdot 7}=\sqrt{63}\]
Pułapka egzaminacyjna: dodawanie pierwiastków
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Nie wolno robić: \(\sqrt{2}+\sqrt{8}=\sqrt{10}\). To błąd.
Najpierw uprość \(\sqrt{8}=2\sqrt{2}\), a dopiero potem dodawaj:
\(\sqrt{2}+\sqrt{8}=\sqrt{2}+2\sqrt{2}=3\sqrt{2}\).
Sprawdź się
- Uprość: \(\sqrt{72}\).
- Oblicz: \(\sqrt{12}\cdot\sqrt{3}\).
- Czy \(2\sqrt{3}+\sqrt{27}\) da się uprościć? Jeśli tak, to jak?
Odpowiedzi: (1) \(\sqrt{72}=\sqrt{36\cdot 2}=6\sqrt{2}\). (2) \(\sqrt{36}=6\). (3) \(\sqrt{27}=3\sqrt{3}\), więc \(2\sqrt{3}+\sqrt{27}=2\sqrt{3}+3\sqrt{3}=5\sqrt{3}\).
