What is (b/M+K/b)(2sqrt(((Kb+Kb))/(bM)))^{-1} ?

The solution to (b/M+K/b)(2sqrt(((Kb+Kb))/(bM)))^{-1} is (b^2+KM)/(2sqrt(2)b\sqrt{M)sqrt(K)}

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simplify (b/M+K/b)(2sqrt(((Kb+Kb))/(bM)))^{-1}

simplify

(\frac{b}{M} + \frac{K}{b}) (2 \frac{( K b + K b )}{b M})^{- 1}

\frac{b ^{2} + K M}{2 2 b M K}

Solution steps

(\frac{b}{M} + \frac{K}{b}) (2 \frac{K b + K b}{b M})^{- 1}

Join

\frac{b}{M} + \frac{K}{b} : \frac{b ^{2} + K M}{b M}

= \frac{b ^{2} + K M}{b M} (2 \frac{K b + K b}{b M})^{- 1}

Simplify

(2 \frac{K b + K b}{b M})^{- 1} : \frac{M}{2 2 K}

= \frac{b ^{2} + K M}{b M} \cdot \frac{M}{2 2 K}

Apply the fraction rule:

\frac{a}{b} \cdot \frac{c}{d} = \frac{a \cdot c}{b \cdot d}

= \frac{( b ^{2} + K M ) M}{b M \cdot 2 2 K}

Apply radical rule:

a = a a

M = M M

= \frac{( b ^{2} + K M ) M}{b M M \cdot 2 2 K}

Cancel the common factor:

M

= \frac{b ^{2} + K M}{b M \cdot 2 2 K}

= \frac{b ^{2} + K M}{2 b M 2 K}

= \frac{b ^{2} + K M}{2 2 b M K}

2 (2) e^{2}

(4 x^{2}) (3 x^{5})

4 \cdot π \cdot (8.85 \cdot 1 0^{- 12}) (8 \cdot 1 0^{- 2})^{2} \cdot 2

e^{- 3 t} cos (3 t)

What is (b/M+K/b)(2sqrt(((Kb+Kb))/(bM)))^{-1} ?
The solution to (b/M+K/b)(2sqrt(((Kb+Kb))/(bM)))^{-1} is (b^2+KM)/(2sqrt(2)b\sqrt{M)sqrt(K)}