Nowa MATeMAtyka 1. Podręcznik. Zakres podstawowy | Strona 39
Zadanie 1. Oblicz.
a)
\[ – 32, – \frac{1}{32},\frac{1}{32}\]
b)
\[ 9, – 27,81\]
c)
\[ 9,\frac{1}{3},\frac{1}{27}\]
d)
\[ 8,8\sqrt{2},\frac{1}{16}\]
Zadanie 2. Zapisz liczbę w postaci 2m, gdzie m jest liczbą całkowitą.
a)
\[ 2^3 \cdot 4^6 = 2^3 \cdot (2^2)^6 = 2^3 \cdot 2^{12} = 2^{3 + 12} = 2^{15}\]
b)
\[ 4^{- 5} \cdot 8^2 = (2^2)^{- 5} \cdot (2^3)^2 = 2^{- 10} \cdot 2^6 = 2^{- 10 + 6} = 2^{- 4}\]
c)
\[ 64^2:32^{- 3} = (2^6)^2:(2^5)^{- 3} = 2^{12}:2^{- 15} = 2^{12 – ( – 15)} = 2^{27}\]
d)
\[ (16^{- 2}:4^{- 8}) \cdot 8^4 = ((2^4)^{- 2}:(2^2)^{- 8}) \cdot (2^3)^4 = (2^{- 8}:2^{- 16}) \cdot 2^{12} = 2^8 \cdot 2^{12} = 2^{20}\]
Zadanie 3. Zapisz podane liczby w postaci potęg o tej samej podstawie.
a)
\[ 2^4,\frac{1}{64} = (\frac{1}{2})^6 = 2^{- 6},8^3 = (2^3)^3 = 2^9,(\sqrt{2})^4 = 2^2,1024^2 = (2^{10})^2 = 2^{20}\]
b)
\[ \frac{1}{81} = (\frac{1}{3})^4 = 3^{- 4},27 = 3^3,9^{- 2} = (3^2)^{- 2} = 3^{- 4},(\sqrt{3})^6 = 3^3,81^{- 5} = (3^4)^{- 5} = 3^{- 20}\]
c)
\[ 0,001 = \frac{1}{1000} = 10^{- 3},100^5 = (10^2)^5 = 10^{10},(\frac{1}{100})^{- 4} = (10^{- 2})^{- 4} = 10^8,(\frac{1}{0,1})^{- 6} = 10^{- 6}\]
d)
\[ 5^{- 3} = (5^2)^{- 3} = 5^{- 6},(\frac{1}{5})^{- 2} = 5^2,(\frac{1}{125})^4 = (5^{- 3})^4 = 5^{- 12},625^{- 5} = (5^4)^{- 5} = 5^{- 20}\]
Zadanie 4. Oblicz.
a)
\[ ( – 2\sqrt{3})^{- 3} = ( – \frac{1}{2\sqrt{3}})^3 = – \frac{1}{8 \cdot 3\sqrt{3}} = – \frac{1}{24\sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{3}} = – \frac{\sqrt{3}}{72}\]
b)
\[(\frac{3}{\sqrt{2}})^{- 4} = (\frac{\sqrt{2}}{3})^4 = \frac{4}{81}\]
c)
\[( – \frac{\sqrt{5}}{10})^{- 3} = ( – \frac{10}{\sqrt{5}})^3 = – \frac{1000}{5\sqrt{5}} \cdot \frac{\sqrt{5}}{\sqrt{5}} = – \frac{1000\sqrt{5}}{25} = – 40\sqrt{5}\]
d)
\[(( – 3\sqrt{2})^{- 1})^{- 2} = ( – 3\sqrt{2})^2 = 9 \cdot 2 = 18\]
Zadanie 5. Oblicz.
a)
\[5^{- 3} \cdot 2^{- 3} = (5 \cdot 2)^{- 3} = 10^{- 3} = 0,001\]
b)
\[0,2^{- 5} \cdot 5^{- 5} = (0,2 \cdot 5)^{- 5} = 1^{- 5} = 1\]
c)
\[8^{- 4} \cdot 4^4 = 8^{- 4} \cdot (\frac{1}{4})^{- 4} = (8 \cdot \frac{1}{4})^{- 4} = 2^{- 4} = \frac{1}{16}\]
d)
\[18^{- 3}:6^{- 3} = (18:6)^{- 3} = 3^{- 3} = \frac{1}{3}^3 = \frac{1}{27}\]
e)
\[(2\frac{3}{7})^5:(\frac{17}{14})^5 = (\frac{17}{7}:\frac{17}{14})^5 = (\frac{17}{7} \cdot \frac{14}{17})^5 = 2^5 =32\]
f)
\[1,3^{- 2}:0,1^{- 2} = ( – \frac{13}{10}:\frac{1}{10})^{- 2} = ( – \frac{13}{10} \cdot \frac{10}{1})^{- 2} = ( – \frac{130}{10})^{- 2} = ( – \frac{1}{13})^2 = \frac{1}{169}\]
g)
\[( – 0,2)^{- 8}:5^8 = ( – \frac{10}{2})^8:5^8 = ( – \frac{10}{2} \cdot \frac{1}{5})^8 = ( – 1)^8 = 1\]
h)
\[( – 1,2)^{- 5}:(\frac{6}{5})^{- 5} = ( – \frac{12}{10}:\frac{6}{5})^{- 5} = ( – \frac{12}{10} \cdot \frac{5}{6})^{- 5} = ( – 1)^{- 5} = – 1\]
Zadanie 6. Oblicz.
a)
\[(\frac{6}{5})^{- 5} \cdot (\frac{7}{6})^{- 3} \cdot (\frac{5}{7})^{- 4} – (\frac{5}{6})^5 \cdot (\frac{6}{7})^3 \cdot (\frac{7}{5})^4 = \frac{3125}{7776} \cdot \frac{216}{343} \cdot \frac{2401}{625} = \frac{5}{36} \cdot \frac{1}{1} \cdot \frac{7}{1} = \frac{35}{36}\]
b)
\[(\frac{2}{21})^{- 3} \cdot 15^{- 3} \cdot 7^{- 3} = (\frac{2}{21} \cdot 15 \cdot 7)^{- 3} = (\frac{210}{21})^{- 3} = (10)^{- 3} = (\frac{1}{10})^3 = \frac{1}{1000}\]
c)
\[0,4^3 \cdot (\frac{5}{16})^3:4^{- 3} = (\frac{4}{10})^3 \cdot (\frac{5}{16})^3:(\frac{1}{4})^3 = (\frac{4}{10} \cdot \frac{5}{16}:\frac{1}{4})^3 = (\frac{4}{10} \cdot \frac{5}{16} \cdot \frac{4}{1})^3 = (\frac{1}{2})^3 = \frac{1}{8}\]
Zadanie 7. Która z liczb jest większa: x czy y?
a)
\[x = 2^4 \cdot 4^{- 2} = 2^4 \cdot (2^2)^{- 2} = 2^4 \cdot 2^{- 4} = 2^0\]
\[y = 4^{- 4}:8^{- 2} = (2^2)^{- 4}:(2^3)^{- 2} = 2^{- 8}:2^{- 6} = 2^{- 2}\]
x > y
b)
\[x = (2^{- 4}:2^{- 6})^{- 1} = (2^2)^{- 1} = 2^{- 2}\]
\[y = (2^{- 4} \cdot 2^{- 3})^{- 1} = (2^{- 7})^{- 1} = 2^7\]
y > x
Zadanie 8. Oblicz.
a)
\[\frac{2^{- 2}}{3^{- 3}} \cdot (\frac{4}{9})^2 = \frac{2^{- 2}}{3^{- 3}} \cdot \frac{16}{81} = \frac{3^3}{2^2} \cdot \frac{2^4}{3^4} = \frac{2^2}{3} = \frac{4}{3}\]
b)
\[((\frac{2}{3})^{- 2})^{- 2} = (\frac{2}{3})^4 = \frac{16}{81}\]
c)
\[\frac{6^0 + 0^6}{6^{- 1}} + (4^6 – 16^3) = \frac{1 + 0}{6^{- 1}} + (4096 – 4096) = \frac{1}{6^{- 1}} = 6\]
d)
\[((\frac{1}{3})^4 \cdot (\frac{2}{3})^{- 5}):6^{- 2} = ((\frac{1}{3})^4 \cdot (\frac{3}{2})^5):(\frac{1}{6})^2 = (\frac{1}{81} \cdot \frac{243}{32}):\frac{1}{36} = \frac{3}{32} \cdot 36 = \frac{108}{32} = 3\frac{3}{8}\]
e)
\[(0,5 \cdot (\frac{1}{8})^{- 6} – 2 \cdot 16^4):7^3 = (\frac{1}{2} \cdot (2^3)^6 – 2 \cdot (2^4)^4):7^3 = (2^{- 1} \cdot 2^{18} – 2 \cdot 2^{16}):7^3 = (2^{17} – 2^{17}):7^3 = 0:7^3 = 0\]
f)
\[((\frac{2}{5})^{- 3}:(\frac{5}{2})^2) \cdot (\frac{5}{2})^{- 4} = ((\frac{5}{2})^3:(\frac{5}{2})^2) \cdot (\frac{5}{2})^{- 4} = (\frac{5}{2})^{3 – 2} \cdot (\frac{5}{2})^{- 4} = (\frac{5}{2})^{1 – 4} = (\frac{5}{2})^{- 3} = (\frac{2}{5})^3 = \frac{8}{125}\]
Zadanie 9. Podaj konieczne założenia i uprość wyrażenie, a następnie oblicz jego wartość dla a = −1/2.
Założenie: a ≠ 0
a)
\[a^3 \cdot a^5 \cdot a^{- 6} = a^{3 + 5 – 6} = a^2 = ( – \frac{1}{2})^2 = \frac{1}{4}\]
b)
\[(a^8 \cdot a^{- 3}):a^2 = a^{(8 – 3) – 2} = a^3 = ( – \frac{1}{2})^3 = – \frac{1}{8}\]
c)
\[(a^4:a^{- 1}) \cdot a^{- 3} = a^{(4 + 1) – 3} = a^2 = ( – \frac{1}{2})^2 = \frac{1}{4}\]
d)
\[(a^7:a^{- 2}):a^{- 4} = a^{(7 + 2) + 4} = a^{13} = ( – \frac{1}{2})^{13} = – \frac{1}{8192}\]
e)
\[(a^{- 1} \cdot a^6)^{- 2} \cdot (a^{- 3})^2 = (a^{( – 1 + 6)})^{- 2} \cdot a^{- 6} = a^{- 10} \cdot a^{- 6} = a^{- 10 – 6} = a^{- 16} = ( – \frac{1}{2})^{- 16} = 2^{16} = 65536\]
f)
\[(a^5:a^{- 4})^2:(a^{- 4})^{- 1} = (a^{(5 + 4)})^2 \cdot a^4 = a^{18}:a^4 = a^{18 – 4} = a^{12} = ( – \frac{1}{2})^{12} = \frac{1}{16384}\]