С решением, пожалуйста!. Created by bogsampa2019. algebra-ru

С решением, пожалуйста!...

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$(\ \underbrace{(\sqrt3\ )^2}_{3}\ )^3=3^3=27\\\\\\(-\underbrace {(-\sqrt{2}\, )^2}_{2}\ )^3=(-2)^3=-8\\\\\\\Big(-\dfrac{1}{5}\, \underbrace {(\sqrt{5})^2}_{5}\Big)^3=\Big(-\dfrac{1}{5}\cdot 5\Big)^3=(-1)^3=-1\\\\\\\Big(-\dfrac{1}{2}\, (\sqrt{3})^{-1}\, \Big)^6=\dfrac{1}{2^6}\cdot \Big(\dfrac{1}{\sqrt{3}}\Big)^6=\dfrac{1}{64}\cdot \dfrac{1}{27}=\dfrac{1}{1728}$

1 votes Thanks 1

arsik32

1) ((√3)^2)^3 = 3^3 = 27

2) (-(-√2)^2)^3 = (-2)^3 = -8

3) (-1/5 * (√5)^2)^3 = (-1/5 * 5)^3 = (-1)^3 = -1

4) (-1/2(√3)^-1)^6 = (-1/2 * 1/√3)^6 = 1/64 * 1/27 = 1/1728

0 votes Thanks 0

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