Question

please help its urgent.

show the whole process step by step .​

Answer 1

Answer:

1. taking LCM
2. taking LCMab+1/b*a/ab+1^m * ab-1/b*a/ab-1^n
3. a/b^m * a/b^n
4. if base same power add
5. a/b^m+n

Answer 2

Step-by-step explanation:

Step 1: Take LCM

In the Numerator we'll get:

((ab + 1)/b)^m × ((ab - 1)/b)^n

= [ ((ab + 1)^m) ÷ (b^m) ] × [ ((ab - 1)^n) ÷ (b^n) ]

In the Denominator, we'll get:

((ab + 1)/a)^m × ((ab - 1)/a)^n

= [ ((ab + 1)^m) ÷ (a^m) ] × [ ((ab - 1)^n) ÷ (a^n) ]

<Step:2 as we can see, the BOLD past of Numerator & Denominator can easily be divided.>

Next Step:

We'll then get:

In the Numerator:

= (1 / b^m) × (1 / b^n)

In the Denominator:

= (1 / a^m) × (1/a^n)

Now, <taking both Numerator & Denominator>

= [ (1 / b^m) × (1 / b^n) ] ÷ [ (1 / a^m) × (1/a^n) ]

= [ (a^m) × (a^n) ] ÷ [ (b^m) × (b^n) ]

= [ a^(m + n) ] ÷ [ b^(m + n) ]

= (a/b)^(m+n)

Hence Proved.

Thanks!