If a- b- c are positive real numbers- then -a-1b-b-1c-c-1a is equal to -abc1-abcabc1

The correct option is D 1 √(a^ 1b)×√(b^ 1c)×√(c^ 1a) = √(b/a)×√(c/b)×√(a/c) = √(1) =1 ...

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Laws of Exponents for Real Numbers

If a, b, c ar...

Question

If a, b, c are positive real numbers, then $\sqrt{a^{- 1} b} \times \sqrt{b^{- 1} c} \times \sqrt{c^{- 1} a}$ is equal to

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abc

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$\sqrt{a b c}$

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$\frac{1}{a b c}$

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Solution

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Similar questions

\sqrt{a^{- 1} b} \times \sqrt{b^{- 1} c} \times \sqrt{c^{- 1} a}

\sqrt{a b c}

\frac{1}{a b c}

\sqrt{a^{- 1} b} . \sqrt{b^{- 1} c} . \sqrt{c^{- 1} a}

a . b

c

a + \frac{1}{b} = 4, b + \frac{1}{c} = 1

c + \frac{1}{a} = \frac{7}{3}

6

a b c

I f s i n^{- 1} a + s i n^{- 1} b + s i n^{- 1} c = π, t h e n t h e v a l u e o f a \sqrt{(1 - a^{2})} + b \sqrt{(1 - b^{2})} + c \sqrt{(1 - c^{2})} w i l l b e

(a + c) \neq 0

∣ ∣ ∣ ∣ \begin{matrix} a & a + 1 & a - 1 - b & b + 1 & b - 1 c & c - 1 & c + 1 \end{matrix} ∣ ∣ ∣ ∣ + ∣ ∣ ∣ ∣ ∣ \begin{matrix} a + 1 & b + 1 & c - 1 a - 1 & b - 1 & c + 1 (- 1)^{n + 2} a & (- 1)^{n + 1} b & (- 1)^{n} c \end{matrix} ∣ ∣ ∣ ∣ ∣ = 0

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