tex]Если а=2, b= -4, c=8 ...

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[tex] \sqrt{(ab) {}^{c} } \times \sqrt{(bc) {}^{a} } \times \sqrt{(ac) {}^{b} } [/tex]
Если а=2, b= -4, c=8

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$\sqrt{(ab)^{c}}\cdot\sqrt{(bc)^{a} } \cdot\sqrt{(ac)^{b}}=\sqrt{a^{c}\cdot b^{c}\cdot b^{a}\cdot c^{a}\cdot a^{b}\cdot c^{b}}=\sqrt{a^{b+c}\cdot b^{a+c}\cdot c^{a+b}}=\\\\=\sqrt{2^{-4+8}\cdot (-4)^{2+8}\cdot 8^{2-4}}=\sqrt{2^{4}\cdot 4^{10}\cdot 8^{-2}}=2^{2}\cdot 4^{5}\cdot \frac{1}{8} =\boxed{512}$

1 votes Thanks 1

sasasihov65 спасибо

lakOR555

Ответ:

512

Объяснение:

$\sqrt{(-4*2)^8} * \sqrt{(-4*8)^2} *\sqrt{(2*8)^{-4}} =\\ \\ = (-4*2)^{8/2} * (4*8)^{2/2} * (2*8)^{-4/2} = \\ \\= (-4*2)^4 * (4*8)^1 * (2*8)^{-2} =\\ \\ =(-8)^4 * 32 * \frac{1}{(2*8)^2} =\\ \\ =4096 * 32 * \frac{1}{16^2} \\ \\ = 4096 * 32 * \frac{1}{256} = \\ \\ =4096 * \frac{1}{256/32} = \\ \\ =4096 * \frac{1}{8} = 512$

2 votes Thanks 1

lakOR555 ++

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