x^2*x^3 = x^5. It is true.

Every rational number has its reciprocal. It is not true.

(x–3)^–4 ≠ x¹², for every rational number x > 0.

Hence, option B is correct.

HCF of two numbers = 28

LCM of two numbers = 336

One number = 112

We have, 

Hence, option C is correct.

The square of 9 = 81

And cubroot of 125 = 5

Therefore, if we divide 81 by 5, the quotient = 16 and the remainder = 1.

Hence, option A is correct.

Let ABC be an equilateral triangle and AD is the height of it. 

Let AB = BC = CA = x; AD = h

AD bisects BC

Hence by Pythagoras theorem:

AD² = AB² – BD²

h² = x² – (x/2)² = x² – x²/4 = 3x²/4

h = √3x/2

Hence ratio of the side and height of an equilateral triangle = x/ h

= x : √3x/2 = 2: √3

Hence, option A is correct.

a, b and c are three natural numbers in ascending order

hence; a<b<c

⇒ (c + a) > b & (c – a) > 0

⇒ c² - a² = (c + a) (c – a) > b

Thus, 10/11 is the least fraction. 
Hence, option B is correct.

3¹ = 3, divided by 5 gives a remainder of 3.

3² = 9, divided by 5 gives a remainder of 4.

3³ = 27, divided by 5 gives a remainder of 2.

3⁴ = 81, divided by 5 gives a remainder of 1. 

…… 

It is clear that the remainder lie between 1 to 4.

Therefore, this process repeats itself.

Now, 

3^{(1+ 4n)} ; remainder = 3

3^{(2+ 4n)} ; remainder = 4

3^{(3+ 4n)} ; remainder = 2

3^{(4+ 4n)} ; remainder = 1 

So, 3⁹⁸ = 3^2+4×24

(4 ×24 = 96)

It ends up being 2+96 = 98, so we have the equation

3^{(2 + 4n)}; remainder = 4

Thus, When the number 3⁹⁸ is divided by 5, the remainder will be 4.

Hence, option D is correct.

Therefore, LCM of 10, 12, 15 and 18 is 2 × 3 × 5 × 2 × 3 = 180

Thus, to find a least perfect number which is divisible by 10, 12, 15 and 18, we will multiply the LCM by 5, i.e, 180 × 5 = 900.

Hence, option D is correct.

Hence, option C is correct.