Question

ans my question I will mark as brainliest ​

Answer 1

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$\huge\mathcal{Your-Answer}$

Answer 2

Answer: 17)Firstly I will prove root 2

(Bear with me as for some reason the equation option is not working so root two will be like this)

Root 2 is rational

So root 2 =p/q (p/q is applicable to all rational numbers in which q not equal to 0)

Now squaring on both sides

(Root 2)²=(p/q)²

2=p²/q²

2q²=p²------(1)

If p² is divisible by 2,then p must be divisible by 2

P=2a(for some integer a)

Substituting the value of p in equation 1

2q²=(2a)²

2q²=4a²

q²=4a²/2

q²=2a²

If q² is divisible by 2 ,then q must be divisible by 2

So p and q must at least have 2 as a common factor

But this contradicts the fact that p and q are co primes.

This contradiction has arisen due to the incorrect assumption that root 2 is rational

Hence it is irrational.

Now let's prove 3 +Root 2

(Note :for this question no need to prove root 2 if small question but 4 marker please do)(this is also for understanding)

3+root2 =p/q

Root2 =p/q - 3

Root2 =p-3q/q

L.H.S is rational so R.H.S should be rational

But this contradicts the fact that root 2 Is irrational.

So our assumption is incorrect .

Hence 3 +Root 2 is irrational

18)Hcf and Icm of 12, 72, 120

The answer is in the below picture

I HOPE THIS helps