Potęgowanie ilorazu klasa 7, proszę o pomoc

Odpowiedź:
Szczegółowe wyjaśnienie:
=(15:5)^10=3^10
=(0,8:0,2)^2=(8/10:2/10)^2=(8/10*10/2)=4^2=16
=(-3:0,7)^2= (-3:7/10)^2=(-3*10/7)^2=(-30/7)^2=(-4 2/7)^2
=(0,6:0,3)^2=(6/10:3/10)^2=(6/10*10/3)=2^2=4
=(ab/2c)^2=a^2b^2/4c^2
=(3x/y)^2=9x^2/y^2
=(7:x)^3=7^3/x^3
=(1/3)^4=1/81
=(10:5)^4=2^4=16
=(-9/3)^3=(-3)^3=-27
=(1/8:1/4)^3=(1/8*4)^3=(1/2)^3=1/8
=2,2^2:1,1^2=(2,2:1,1)^2=(22/10:11/10)^2=(22/10*10/11)^2=2^2=4
=(3z)^2=9z^2
=8/125
=(1 1/4)^2=(5/4)^2=25/16=1 9/16
=(2,5:0,5)^3=(25/10:5/10)^3=(25/10*10/5)^3=563=125
=(-6k/2)^2= 36k^2/4=9k^2
=x^3y^/27
=a^2/36
[tex]15^{10}:5^{10} = (15:5})^{10} =3^{10}=59049\\\\(0,8)^{2}:(0,2)^{2} = (0,8:0,2)^{2} = 4^{2} = 16\\\\(-3)^{2}:(0,7)^{2} = (-3:\frac{7}{10})^{2} = (-3\cdot\frac{10}{7})^{2} }=(-\frac{30}{7})^{2}=\frac{900}{49} = 18\frac{18}{49}\\\\(06)^{2}:(0,3)^{2} = (0,6:0,3)^{2} =2^{2} = 4[/tex]
[tex](\frac{ab}{2c})^{2} = \frac{a^{2}b^{2}}{4c^{2}}\\\\(\frac{3x}{y})^{2} = \frac{9x^{2}}{y^{2}}\\\\(7:x)^{3} = \frac{7^{3}}{x^{3}} = \frac{343}{x^{3}} \ \ \ (zal: x \neq 0)[/tex]
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[tex](9z:3)^{2} = (3z)^{2} = 9z^{2}\\\\(\frac{2}{5})^{3}=\frac{2^{3}}{5^{3}} =\frac{8}{125}\\\\(1\frac{1}{4})^{2} = (\frac{5}{4})^{2} = \frac{5^{2}}{4^{2}} = \frac{25}{16} = 1\frac{9}{16}\\\\(2,5)^{3}:(0,5)^{3} = (2,5:0,5)^{3} = 5^{3} = 125\\\\(-6k)^{2}:2^{2} = (-6k:2)^{2} = (-3k)^{2} = 9k^{2}[/tex]
[tex](\frac{xy}{3})^{3} = \frac{x^{3}y^{3}}{3^{3}} = \frac{x^{3}y^{3}}{27}\\\\(\frac{a}{6})^{2} = \frac{a^{2}}{6^{2}} = \frac{a^{2}}{36}[/tex]
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[tex](\frac{1}{3})^{4} = \frac{1^{4}}{3^{4}} = \frac{1}{81}\\\\10^{4}:5^{4} = (10:5)^{4} = 2^{4} = 16\\\\9^{3}:(-3)^{3} = (9:(-3))^{3} = (-3)^{3} = -27\\\\(\frac{1}{8})^{3}:(\frac{1}{4})^{3} =(\frac{1}{8}:\frac{1}{4})^{3} = (\frac{1}{8}\cdot4)^{3} = (\frac{1}{2})^{3}= \frac{1}{8}\\\\(2,2)^{2}:(1,1)^{2} = (2,2:1,1)^{2} = 2^{2} = 4[/tex]