What is 2(2A)^{-1}(B^{-1}B)^{-1}AC(DC)^{-1}((D^{-1})^{-1})^2 ?

The solution to 2(2A)^{-1}(B^{-1}B)^{-1}AC(DC)^{-1}((D^{-1})^{-1})^2 is D

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simplify 2(2A)^{-1}(B^{-1}B)^{-1}AC(DC)^{-1}((D^{-1})^{-1})^2

simplify

2 (2 A)^{- 1} (B^{- 1} B)^{- 1} A C (D C)^{- 1} ((D^{- 1})^{- 1})^{2}

D

Solution steps

2 (2 A)^{- 1} (B^{- 1} B)^{- 1} A C (D C)^{- 1} ((D^{- 1})^{- 1})^{2}

Apply exponent rule:

(a \cdot b)^{n} = a^{n} b^{n}

(2 A)^{- 1} = 2^{- 1} A^{- 1}

= 2^{- 1} \cdot 2 A^{- 1} A C ((D^{- 1})^{- 1})^{2} (B^{- 1} B)^{- 1} (C D)^{- 1}

Apply exponent rule:

a^{b} \cdot a^{c} = a^{b + c}

A^{- 1} A = A^{- 1 + 1}

= 2 \cdot 2^{- 1} (B^{- 1} B)^{- 1} A^{- 1 + 1} C (D C)^{- 1} ((D^{- 1})^{- 1})^{2}

Add/Subtract the numbers:

- 1 + 1 = 0

= 2 \cdot 2^{- 1} (B^{- 1} B)^{- 1} A^{0} C (D C)^{- 1} ((D^{- 1})^{- 1})^{2}

Apply rule

a^{0} = 1, a \neq = 0

A^{0} = 1

= 2^{- 1} \cdot 2 \cdot 1 \cdot C ((D^{- 1})^{- 1})^{2} (B^{- 1} B)^{- 1} (C D)^{- 1}

Apply exponent rule:

a^{m} b^{m} = (ab)^{m},

assuming

a \geq 0, b \geq 0

(B^{- 1} B)^{- 1} (C D)^{- 1} = (B^{- 1} BC D)^{- 1}

= 2^{- 1} \cdot 2 \cdot 1 \cdot C ((D^{- 1})^{- 1})^{2} (B^{- 1} BC D)^{- 1}

Simplify

B^{- 1} B D C : C D

= 2^{- 1} \cdot 2 \cdot 1 \cdot C ((D^{- 1})^{- 1})^{2} (C D)^{- 1}

Apply exponent rule:

(ab)^{c} = a^{c} b^{c}

(C D)^{- 1} = C^{- 1} D^{- 1}

= 2 \cdot 2^{- 1} C^{- 1} D^{- 1} \cdot 1 \cdot C ((D^{- 1})^{- 1})^{2}

Apply exponent rule:

a^{b} \cdot a^{c} = a^{b + c}

2 \cdot 2^{- 1} = 2^{1 - 1}

= 2^{1 - 1} C^{- 1} D^{- 1} \cdot 1 \cdot C ((D^{- 1})^{- 1})^{2}

Subtract the numbers:

1 - 1 = 0

= 2^{0} C^{- 1} D^{- 1} \cdot 1 \cdot C ((D^{- 1})^{- 1})^{2}

Apply exponent rule:

a^{b} \cdot a^{c} = a^{b + c}

C^{- 1} C = C^{- 1 + 1}

= 2^{0} D^{- 1} \cdot 1 \cdot C^{- 1 + 1} ((D^{- 1})^{- 1})^{2}

Add/Subtract the numbers:

- 1 + 1 = 0

= 2^{0} D^{- 1} \cdot 1 \cdot C^{0} ((D^{- 1})^{- 1})^{2}

Apply rule

a^{0} = 1, a \neq = 0

= 1 \cdot 1 \cdot 1 \cdot D^{- 1} ((D^{- 1})^{- 1})^{2}

((D^{- 1})^{- 1})^{2} = D^{2}

= 1 \cdot 1 \cdot 1 \cdot D^{- 1} D^{2}

Apply exponent rule:

a^{- 1} = \frac{1}{a}

D^{- 1} = \frac{1}{D}

= 1 \cdot 1 \cdot 1 \cdot D^{2} \frac{1}{D}

Multiply fractions:

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

= 1 \cdot 1 \cdot 1 \cdot \frac{1 \cdot D ^{2}}{D}

Simplify

\frac{1 \cdot D ^{2}}{D} : D

= 1 \cdot 1 \cdot 1 \cdot D

Multiply:

1 \cdot 1 \cdot 1 \cdot D = D

= D

\frac{( n + 1 ! ) ^{2}}{( n ! ) ^{2}}

(41)^{2}

\frac{x ^{a - 1}}{y ^{a - 1}}

5 e^{- 3 x}

What is 2(2A)^{-1}(B^{-1}B)^{-1}AC(DC)^{-1}((D^{-1})^{-1})^2 ?
The solution to 2(2A)^{-1}(B^{-1}B)^{-1}AC(DC)^{-1}((D^{-1})^{-1})^2 is D