Question

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Answer 1

Step-by-step explanation:

Suppose 3–√=ab, where a and b are co-prime integers, b≠0

Squaring both sides,

⇒3= a2b2

Multiplying with b on both sides,

⇒3b= a2b

LHS = 3×b =Integer

RHS = a2b= Integer Integer =Rational Number

⇒LHS≠RHS

∴ Our supposition is wrong.

⇒3–√ is irrational.

Suppose 15+173–√ is a rational number.

∴15+173–√=ab, where a and b are co-prime, b≠0

⇒173–√=ab−15

3–√=a−15b17b

a−15b17b is rational number,

3–√ is irrational.

∴3–√≠a−15b17b

∴ Our supposition is wrong.

⇒15+173–√ is irrational.

Answer 2

Answer:

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Step-by-step explanation:

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