Nowa MATeMAtyka 1. Podręcznik. Zakres podstawowy | Strona 29
Ćwiczenie 4. Oblicz.
a)
\[\sqrt{4 \cdot 81} = \sqrt{324} = 18\]
\[\sqrt{25 \cdot 0,36} = \sqrt{9} = 3\]
\[\sqrt{0,09 \cdot 361} = \sqrt{32,49} = 5,7\]
b)
\[\sqrt{\frac{121}{144}} = \frac{11}{12}\]
\[\sqrt{\frac{361}{400}} = \frac{19}{20}\]
\[\sqrt{\frac{576}{625}} = \frac{24}{25}\]
c)
\[\sqrt{2} \cdot \sqrt{8} = \sqrt{2 \cdot 8} = \sqrt{16} = 4\]
\[\sqrt{6} \cdot \sqrt{1,5} = \sqrt{6 \cdot 1,5} = \sqrt{9} = 3\]
\[\sqrt{2} \cdot \sqrt{5} \cdot \sqrt{10} = \sqrt{2 \cdot 5 \cdot 10} = \sqrt{100} = 10\]
d)
\[\frac{\sqrt{54}}{\sqrt{6}} = \sqrt{9} = 3\]
\[\frac{\sqrt{3}}{\sqrt{12}} = \sqrt{\frac{1}{4}} = \frac{1}{2}\]
\[\frac{\sqrt{10} \cdot \sqrt{14}}{\sqrt{35}} = \sqrt{\frac{140}{35}} = \sqrt{4} = 2\]
Zadanie 1. Oblicz.
a)
\[\sqrt{121} + \sqrt{49} – \sqrt{225} = 11 + 7 – 15 = 3\]
b)
\[\sqrt{196} – \sqrt{169} – \sqrt{144} = 14 – 13 – 12 = – 11\]
c)
\[\sqrt{0,25} + \sqrt{1,44} + \sqrt{6,25} = 0,5 + 1,2 + 2,5 = 4,2\]
d)
\[\sqrt{3,61} – \sqrt{1,21} – \sqrt{0,09} = 1,9 – 1,1 – 0,3 = 0,5\]
e)
\[\sqrt{\frac{81}{400}} + \sqrt{\frac{9}{100}} – \sqrt{\frac{64}{25}} = \frac{9}{20} + \frac{3}{10} – \frac{8}{5} = \frac{9}{20} + \frac{6}{20} – \frac{32}{20} = – \frac{17}{20}\]
f)
\[\sqrt{3\frac{6}{25}} – \sqrt{2\frac{1}{4}} + \sqrt{1\frac{7}{9}} = \sqrt{\frac{81}{25}} – \sqrt{\frac{9}{4}} + \sqrt{\frac{16}{9}} = \frac{9}{5} – \frac{3}{2} + \frac{4}{3} = \frac{54}{30} – \frac{45}{30} + \frac{40}{30} = \frac{49}{30} = 1\frac{19}{30}\]
Zadanie 2. Uzasadnij, że:
a)
\[\sqrt{1\frac{9}{16}} \neq \sqrt{1} + \sqrt{\frac{9}{16}}\]
\[\sqrt{\frac{25}{16}} \neq 1 + \frac{3}{4}\]
\[\frac{5}{4} \neq \frac{7}{4}\]
b)
\[\sqrt{2\frac{1}{4}} \neq \sqrt{2} + \sqrt{\frac{1}{4}}\]
\[\sqrt{2\frac{1}{4}} \neq \sqrt{2} + \sqrt{\frac{1}{4}}\]
√2 ≈ 1,41 + 0,5 ≈ 1,91 zatem
\[\ 1,4 \neq ( \approx 1,91)\]
c)
\[\sqrt{4\frac{1}{9}} \neq \sqrt{4} + \sqrt{\frac{1}{9}}\]
\[\sqrt{\frac{37}{9}} \neq 2 + \frac{1}{3}\]
\[( \approx \frac{6,08}{3}) \neq 2,(3)\]
\[( \approx \frac{6,08}{3}) \neq 2,(3)\]
\[( \approx 2,08 \neq 2,(3)\]
Zadanie 3. Podaj wszystkie liczby naturalne leżące na osi między liczbami:
a)
√2 ≈ 1,41
√33 ≈ 5,74
Odpowiedź: 2, 3, 4, 5
a)
√10 ≈ 3,16
√140 ≈ 11,83
Odpowiedź: 4, 5, 6, 7, 8, 9, 10, 11
c)
√80 ≈ 8,94
√300 ≈ 17,32
Odpowiedź: 9, 10, 11, 12, 13, 14, 15, 16, 17
Zadanie 4. Wyłącz czynnik przed pierwiastek.
a)
\[\sqrt{18} = \sqrt{9 \cdot 2} = 3\sqrt{2}\]
b)
\[\sqrt{24} = \sqrt{4 \cdot 6} = 2\sqrt{6}\]
c)
\[\sqrt{48} = \sqrt{16 \cdot 3} = 4\sqrt{3}\]
d)
\[\sqrt{96} = \sqrt{16 \cdot 6} = 4\sqrt{6}\]
e)
\[\sqrt{108} = \sqrt{36 \cdot 3} = 6\sqrt{3}\]
f)
\[\sqrt{252} = \sqrt{36 \cdot 7} = 6\sqrt{7}\]
g)
\[\sqrt{392} = \sqrt{196 \cdot 2} = 14\sqrt{2}\]
h)
\[\sqrt{450} = \sqrt{225 \cdot 2} = 15\sqrt{2}\]
Zadanie 5. Zapisz liczbę w postaci a√2.
a)
\[4\sqrt{2} + \sqrt{8} = 4\sqrt{2} + \sqrt{4 \cdot 2} = 4\sqrt{2} + 2\sqrt{2} = 6\sqrt{2}\]
b)
\[\sqrt{32} – 3\sqrt{2} = \sqrt{16 \cdot 2} – 3\sqrt{2} = 4\sqrt{2} – 3\sqrt{2} = \sqrt{2}\]
c)
\[\sqrt{18} + \sqrt{98} = \sqrt{9 \cdot 2} + \sqrt{49 \cdot 2} = 3\sqrt{2} + 7\sqrt{2} = 10\sqrt{2}\]
d)
\[\sqrt{200} – \sqrt{50} = \sqrt{100 \cdot 2} – \sqrt{25 \cdot 2} = 10\sqrt{2} – 5\sqrt{2} = 5\sqrt{2}\]
e)
\[\sqrt{18} + \sqrt{72} + \sqrt{242} = \sqrt{9 \cdot 2} + \sqrt{36 \cdot 2} + \sqrt{121 \cdot 2} = 3\sqrt{2} + 6\sqrt{2} + 11\sqrt{2} = 20\sqrt{2}\]
f)
\[\sqrt{800} + \sqrt{242} – \sqrt{162} = \sqrt{400 \cdot 2} + \sqrt{121 \cdot 2} – \sqrt{81 \cdot 2} = 20\sqrt{2} + 11\sqrt{2} – 9\sqrt{2} = 22\sqrt{2}\]
Zadanie 6. Zapisz liczbę w postaci a√b.
a)
\[7\sqrt{5} + \sqrt{20} = 7\sqrt{5} + \sqrt{4 \cdot 5} = 7\sqrt{5} + 2\sqrt{5} = 9\sqrt{5}\]
b)
\[\sqrt{48} – \sqrt{27} = \sqrt{16 \cdot 3} – \sqrt{9 \cdot 3} = 4\sqrt{3} – 3\sqrt{3} = \sqrt{3}\]
c)
\[\sqrt{12} + \sqrt{75} = \sqrt{4 \cdot 3} + \sqrt{25 \cdot 3} = 2\sqrt{3} + 5\sqrt{3} = 7\sqrt{3}\]
d)
\[\sqrt{45} – \sqrt{125} = \sqrt{9 \cdot 5} – \sqrt{25 \cdot 5} = 3\sqrt{5} – 5\sqrt{5} = – 2\sqrt{5}\]
e)
\[0,2\sqrt{50} + 0,8\sqrt{72} – 0,3\sqrt{32} = 0,2\sqrt{25 \cdot 2} + 0,8\sqrt{36 \cdot 2} – 0,3\sqrt{16 \cdot 2} = 0,2 \cdot 5\sqrt{2} + 0,8 \cdot 6\sqrt{2} – 0,3 \cdot 4\sqrt{2} = 1\sqrt{2} + 4,8\sqrt{2} – 1,2\sqrt{2} = 4,6\sqrt{2}\]
f)
\[3\sqrt{20} – \frac{1}{3}\sqrt{45} – 5\sqrt{180} = 3\sqrt{4 \cdot 5} – \frac{1}{3}\sqrt{9 \cdot 5} – 5\sqrt{36 \cdot 5} = 3 \cdot 2\sqrt{5} – \frac{1}{3} \cdot 3\sqrt{5} – 5 \cdot 6\sqrt{5} = 6\sqrt{5} – \sqrt{5} – 30\sqrt{5} = – 25\sqrt{5}\]