DawaLama DawaLama
  • 24-05-2019
  • Mathematics
contestada

Solve this with processes and teach me clearly.​

Solve this with processes and teach me clearly class=

Respuesta :

t51817361
t51817361 t51817361
  • 24-05-2019

Answer: one.

Step-by-step explanation:

Let's break down the numerator first.

That ugly mess can be re-written as:

2^(2x+2y) times 2^(2y+2z) times 2^(2z+2x).    Recall that (a^b)+(a^c)=a^(b+c).

2^(2x+2y) times 2^(2y+2z) times 2^(2z+2x) equals 2^(4x+4y+4z).

Let's break down the denominator now.

That mess can be re-written as:

2^x times 2^y times 2^z.

That can be re-written as 2^(x+y+z).

2^(x+y+z) all to the 4th power equals 2^(4x+4y+4z).

The denominator and numerator equal one another.

Bam.

Ask if you have any questions.

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