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324 = 252 x 1 + 72

252 = 72 x 3+ 36

72 = 36 x 2 + 0

HCF(324, 25) = 36

180 = 36 x 5 + 0

HCF (36, 180) = 36

:. HCF of 180, 252 and 324 is 36.

(b) 180 

It is given that: 

a = (2² × 3³ × 5⁴ ) and b = (2³ × 3² × 5) 

∴HCF (a,b) = Product of smallest power of each common prime factor in the numbers. 

 = 2² × 3² × 5 

 = 180

(c) 60 

HCF = (2³ × 3² × 5, 2² × 3³ × 5² , 2⁴ × 3 × 5³ × 7) 

HCF = Product of smallest power of each common prime factor in the numbers 

= 2² × 3 × 5 

 = 60

(i) Yes (5 – √3) + (5 + √3) = 10, a rational number

(ii) Yes \((5+\sqrt[3]{7})-(-6+\sqrt[3]{7})=11\), a rational number

(iii) Yes (5 + √3) (5 – √3) = 25 – 3 = 22, a rational number

(iv) Yes \(\frac{5\sqrt{3}}{\sqrt{3}}=5\), a rational number

(i) 43.123456789

Since 43.123456789 has terminating decimal expansion. Hence, its denominator is of the form 2^m x 5ⁿ, where m, n are non-negative integers.

(ii) \(43.\overline{123456789}\)

Since the given rational has non-terminating decimal expansion. So, its denominator has factors other than 2 or 5.

(iii) \(27.\overline{142857}\)

Since the given rational number has non-terminating decimal expansion. So, its denominator has factors other than 2 or 5.

(iv) 0.120120012000120000….

Since 0.120120012000120000…. has non-terminating decimal expansion. So, its denominator has factors other than 2 or 5.

(i) 43.123456789

\(\frac{43123456789}{100000000}\) = \(\frac{43123456789}{2^9\times 5^9}\)

Since Prime factorization of the denominator is in the form 2^m x 5ⁿ, where m, n are non -negative integers.

(ii) Since \(43.\overline{123456789}\) has non - terminating repeating decimal expression, therefore

Prime factorization of the denominator contains factors other than 2 or 5.

(iii) Since \(27.\overline{142857}\) has non - terminating repeating decimal expression, therefore

Prime factorization of the denominator contains factors other than 2 or 5.

(iv) Since \(27.\overline{142857}\) has non - terminating repeating decimal expression, therefore

Prime factorization of the denominator contains factors other than 2 or 5.

Let’s assume on the contrary that \(\frac{1}{\sqrt2}\) is a rational number. Then, there exist positive integers a and b such that 

\(\frac{1}{\sqrt2}\) = \(\frac{a}{b}\) where, a and b, are co-primes 

⇒ (\(\frac{1}{\sqrt2}\))² = (\(\frac{a}{b}\))² 

⇒ \(\frac{1}{2}\) = \(\frac{a^2}{b^2}\) 

⇒ 2a² = b² 

⇒ 2 | b² [∵ 2|2b² and 2a² = b²] 

⇒ 2 | b ………… (ii) 

⇒ b = 2c for some integer c. 

⇒ b² = 4c² 

⇒ 2a² = 4c² [∵ b² = 2a²] 

⇒ a² = 2c² 

⇒ 2 | a² ⇒ 2 | a ……… (i) 

From (i) and (ii), we can infer that 2 is a common factor of a and b. But, this contradicts the fact that a and b are co-primes. So, our assumption is incorrect. 

Hence, \(\frac{1}{\sqrt2}\) is an irrational number.

Let’s assume on the contrary that 5√2 is a rational number. 

Then, there exist positive integers a and b such that 

5√2 = \(\frac{a}{b}\) where, a and b, are co-primes 

⇒ √2 = \(\frac{a}{5b}\) 

⇒ √2 is rational [∵ a and b are integers ∴ \(\frac{a}{5b}\) is a rational number] 

This contradicts the fact that √2 is irrational. So, our assumption is incorrect. 

Hence, 5√2 is an irrational number.

Quantity of oil A = 120 liters
Quantity of oil B = 180liters
Quantity of oil C = 240liters
We want to fill oils A, B and C in tins of the same capacity

∴ The greatest capacity of the tin chat can hold oil. A, B and C = HCF of 120, 180 and 240
By fundamental theorem of arithmetic
120 = 2³ × 3 × 5
180 = 2² × 3² × 5
240 = 2⁴ × 3 × 5
HCF = 2² × 3 × 5 = 4 × 3 × 5 = 60 litres
The greatest capacity of tin = 60 litres

Let’s assume on the contrary that 4 + √2 is a rational number. 

Then, there exist co prime positive integers a and b such that 

4 + √2 = \(\frac{a}{b}\) 

⇒ √2 = \(\frac{a}{b}\) + 4 

⇒ √2 = \(\frac{(a + 4b)}{b}\)

⇒ √2 is rational [∵ a and b are integers ∴ \(\frac{(a + 4b)}{b}\) is a rational number] 

This contradicts the fact that √2 is irrational. So, our assumption is incorrect. 

Hence, 4 + √2 is an irrational number.

To find greatest capacity of tin we should find HCF of 120 and 180 and 240

Prime factors of 120 = 2 × 2 × 2 × 3 × 5

Prime factors of 180 = 2 × 2 × 3 × 3 × 5

Prime factors of 240 = 2 × 2 × 2 × 2 × 3 × 5

Therefore HCF of 120, 180 and 240 is:

2 × 2 × 3 × 5 = 60

Therefore the greatest capacity of a tin is 60 Liters

Let’s assume on the contrary that \(\frac{3}{(2√5)}\) is a rational number. Then, there exist co – prime positive integers a and b such that 

\(\frac{3}{(2√5)}\)= \(\frac{a}{b}\)

⇒ √5 = \(\frac{3b}{2a}\)

⇒ √5 is rational [∵ 3, 2, a and b are integers ∴ \(\frac{3b}{2a}\) is a rational number] 

This contradicts the fact that √5 is irrational. So, our assumption is incorrect.

Hence, \(\frac{3}{(2√5)}\) is an irrational number.

Let us assume, to the contrary, that 3√2 is rational.

That is, we can find coprime a and b (b≠0) such that 3√2=a /b 

Rearranging, we get √2=a/3b

Since 3, a and b are integers, a/3b is rational, and so √2 is rational.

But this contradicts the fact that √2 is irrational.

So, we conclude that 3√2 is irrational.

Let (4+3√2) be a rational number. 

Then both (4+3√2) and 4 are rational. 

⇒ (4+3√2 – 4) = 3√2 = rational [∵Difference of two rational numbers is rational] 

⇒ 3√2 is rational. 

⇒ 1/3 (3√2) is rational. [∵ Product of two rational numbers is rational] 

⇒ √2 is rational. 

This contradicts the fact that √2 is irrational (when 2 is prime, √2 is irrational) 

Hence, (4 + 3√2 ) is irrational.

Let, 5 + 3√2 be rational. 

Hence, 5 and 5 + 3√2 are rational. 

∴ (5 + 3√2 – 5) = 3√2 = rational [∵Difference of two rational is rational] 

∴ 1 3 × 3√2 = √2 = rational [∵Product of two rational is rational] 

This contradicts the fact that √2 is irrational. 

The contradiction arises by assuming 5 + 3√2 is rational. 

Hence, 5 + 3√2 is irrational.

Let’s assume on the contrary that 6+√2 is a rational number. 

Then, there exist co prime positive integers a and b such that 

6 + √2 = \(\frac{a}{b}\) 

⇒ √2 = \(\frac{a}{b – 6}\) 

⇒ √2 = \(\frac{(a-6b)}{b }\)

⇒ √2 is rational [∵ a and b are integers ∴ \(\frac{(a-6b)}{b }\)is a rational number] 

This contradicts the fact that √2 is irrational. So, our assumption is incorrect. 

Hence, 6 + √2 is an irrational number.

Let us assume that 3√7 is rational

Here 3 is rational and √7 is rational.

As we know, Product of two rational numbers is rational.

But √7 is rational, which is contradictory to our assumption.

So, 3√7 is irrational

To find maximum number of columns we should find HCF of 616 and 32

Using Euclid’s algorithms:

Let a = 616 and b = 32

a = bq + r, (o ≤ r ≤ b)

616 = 32×19+8

32 = 8×4+0

∴ HCF of 616 and 32 is 8

Therefore the maximum number of columns in which army contingent to march is 8

Let’s assume on the contrary that \(\frac{2}{\sqrt7}\) is a rational number. Then, there exist co-prime positive integers a and b such that 

\(\frac{2}{\sqrt7}\) = \(\frac{a}{b}\)

⇒ √7 =\(\frac{2b}{a}\)

⇒ √7 is rational [∵ 2, a and b are integers ∴ \(\frac{2b}{a}\) is a rational number] 

This contradicts the fact that √7 is irrational. So, our assumption is incorrect. 

Hence, \(\frac{2}{\sqrt7}\) is an irrational number.

As per Euclid’s division lemma: For any two positive integers, say a and b, there exit unique integers q and r, such that a = bq + r; where 0 ≤ r < b.

Also written as: Dividend = (Divisor x Quotient) + Remainder

Let us assume that 5 + √7 is a rational number. 

So, we can find co-prime integers ‘a’ and ‘b’ (b ≠ 0) such that

 \(5+\sqrt7 = \frac{a}{b}\) 

\(\therefore \sqrt7 = \frac{a}{b} - 5\) 

Since, ‘a’ and ‘b’ are integers,\( \sqrt[a]b \) – 5 is a rational number and so √7 is a rational number. 

∴ But this contradicts the fact that √7 is an irrational number. 

Our assumption that 5 + √7 is a rational number is wrong. 

∴ 5 + √7 is an irrational number.

Let us assume that (3 + √2 ) is rational.

Subtract 3 form given number, considering 3 is a rational number.

As we know, Difference of two rational numbers is a rational.

(3 + √2 ) – 3 is rational

√2 is rational

Which is contradictory to our assumption.

So, (3 + √2 ) is irrational

Let us assume that (5 + 3√2 ) is rational.

Subtract 5 form given number, considering 5 is a rational number.

As we know, Difference of two rational numbers is a rational.

(5 + 3√2 ) – 5 is rational

3√2 is rational

And, 3 is rational and √2 is rational.

(Product of two rational numbers is rational)

√2 is rational

Which is contradictory to our assumption.

So, (5 + 3√2 ) is irrational

\(\frac{2}{\sqrt7}\) = \(\frac{2}{\sqrt7}\) x \(\frac{\sqrt7}{\sqrt7}\) = \(\frac{2}{7}\)\(\sqrt7\)

Let  \(\frac{2}{7}\)\(\sqrt7\) is a rational number.

∴ \(\frac{2}{7}\)\(\sqrt7\) = \(\frac{p}{q}\), where p and q are some integers and HCF(p,q) = 1 ….(1)

⇒ 2√7q = 7p 

⇒(2√7q)² = (7p)² 

⇒7(4q² ) = 49p² 

⇒4q² = 7p² 

⇒ q² is divisible by 7 

⇒ q is divisible by 7 …..(2) 

Let q = 7m, where m is some integer. 

∴2√7q = 7p 

⇒ [2√7 (7m)]² = (7p)² 

⇒343(4m² ) = 49p² 

⇒ 7(4m² ) = p² 

⇒ p² is divisible by 7 

⇒ p is divisible by 7 ….(3) 

From (2) and (3), 7 is a common factor of both p and q, which contradicts (1). 

Hence, our assumption is wrong.

Thus, \(\frac{2}{\sqrt7}\) is irrational.

Let us assume that 2/√7 is a rational number.

2/√7 x √7/√7 = 2√7/7 is a rational number

Which is only possible if 2/7 is rational and √7 is rational.

But the fact is √7 is an irrational.

Which is contradicts to our assumption.

2/√7 is an irrational number. Hence proved.

Let us assume that 2 – 3√5 is rational.

Subtract given number form 2, considering 2 is a rational number.

As we know, Difference of two rational numbers is a rational.

2 – (2 – 3√5) is rational

3√5 is rational

Above result is possible if 3 is rational and √5 is rational.

Because, product of two rational numbers is rational

But the fact is √5 is an irrational

Our assumption is wrong, and

(2 – 3√5) is irrational

Let’s assume on the contrary that √5 + √3 is a rational number. Then, there exist co prime positive integers a and b such that 

√5 + √3 = \(\frac{a}{b}\)

⇒ √5 = (\(\frac{a}{b}\)) – √3

⇒ (√5)2 = ((\(\frac{a}{b}\)) – √3)² [Squaring on both sides] 

⇒ 5 = (\(\frac{a^2}{b^2}\)) + 3 – (2√3\(\frac{a}{b}\)) 

⇒ (\(\frac{a^2}{b^2}\)) – 2 = (2√3\(\frac{a}{b}\)) 

⇒ (\(\frac{a}{b}\)) – (2\(\frac{b}{a}\)) = 2√3 

⇒ \(\frac{(a^2 – 2b^2)}{2ab}\) = √3 

⇒ √3 is rational [∵ a and b are integers ∴ \(\frac{(a^2 – 2b^2)}{2ab}\) is rational] 

This contradicts the fact that √3 is irrational. So, our assumption is incorrect.

Hence, √5 + √3 is an irrational number.

Let us assume that (2 – √3) is a rational.

Subtract given number form 2, considering 2 is a rational number.

As we know, Difference of two rational numbers is a rational.

So, 2 – (2 – √3 ) is rational

√3 is rational

Which is contradictory.

Thus, (2 – √3) is an irrational.

Let’s assume on the contrary that 5 – 2√3 is a rational number.

Then, there exist co prime positive integers a and b such that 

5 – 2√3 = \(\frac{a}{b}\) 

⇒ 2√3 = 5 – \(\frac{a}{b}\) 

⇒ √2 = \(\frac{(5b – a)}{(2b)}\) 

⇒ √2 is rational [∵ 2, a and b are integers ∴ \(\frac{(5b – a)}{(2b)}\) is a rational number] 

This contradicts the fact that √2 is irrational. So, our assumption is incorrect. 

Hence, 5 – 2√3 is an irrational number.

Let’s assume on the contrary that 4 – 5√2 is a rational number. 

Then, there exist co prime positive integers a and b such that 

4 – 5√2 = \(\frac{a}{b}\)

⇒ 5√2 = 4 – \(\frac{a}{b}\) 

⇒ √2 = \(\frac{(4b – a)}{(5b)}\)

⇒ √2 is rational [∵ 5, a and b are integers ∴ \(\frac{(4b – a)}{(5b)}\) is a rational number] 

This contradicts the fact that √2 is irrational. So, our assumption is incorrect. 

Hence, 4 – 5√2 is an irrational number.

Let’s assume on the contrary that 2 – 3√5 is a rational number. 

Then, there exist co prime positive integers a and b such that 

2 – 3√5 = \(\frac{a}{b}\)

⇒ 3√5 = 2 – \(\frac{a}{b}\)

⇒ √5 = \(\frac{(2b – a)}{(3b)}\)

⇒ √5 is rational [∵ 3, a and b are integers ∴ \(\frac{(2b – a)}{(3b)}\) is a rational number] 

This contradicts the fact that √5 is irrational. So, our assumption is incorrect. 

Hence, 2 – 3√5 is an irrational number.

Let 5√2 is a rational number.

∴ 5√2 = \(\frac{p}{q}\), where p and q are some integers and HCF(p, q) = 1 …(1)

⇒5√2q = p 

⇒(5√2q)² = p² 

⇒ 2(25q²) = p² 

⇒ p² is divisible by 2 

⇒ p is divisible by 2 ….(2) 

Let p = 2m, where m is some integer. 

∴5√2q = 2m 

⇒(5√2q)² = (2m)² 

⇒2(25q²) = 4m² 

⇒25q² = 2m² 

⇒ q² is divisible by 2 

⇒ q is divisible by 2 ….(3) 

From (2) and (3) is a common factor of both p and q, which contradicts (1). 

Hence, our assumption is wrong. 

Thus, 5√2 is irrational.

Let us assume that (4 – 5√2 ) is a rational.

Subtract given number form 4, considering 4 is a rational number.

As we know, Difference of two rational numbers is a rational.

4 – (4 – 5√2 ) is rational

5√2 is rational

Which is only possible if 5 is rational and √2 is rational

As we know, product of two rational number is rational.

But the fact is √2 is an irrational.

Which is contradict to our assumption.

Hence, 4 – 5√2 is irrational. Hence Proved.

Let’s assume on the contrary that 2√3 – 1 is a rational number. 

Then, there exist co prime positive integers a and b such that 

2√3 – 1 = \(\frac{a}{b}\) 

⇒ 2√3 = \(\frac{a}{b}\) – 1 

⇒ √3 = \(\frac{(a – b)}{(2b)}\)

⇒ √3 is rational [∵ 2, a and b are integers ∴ \(\frac{(a – b)}{(2b)}\) is a rational number] 

This contradicts the fact that √3 is irrational. So, our assumption is incorrect.

Hence, 2√3 – 1 is an irrational number.

Let us assume that 5√2 is a rational.

Which is only possible if 5 is rational and √2 is rational

As we know, product of two rational number is rational.

But the fact is √2 is an irrational.

Which is contradict to our assumption.

5√2 is an irrational. Hence proved.

Let x = 5 - 2√3 be a rational number. 

x = 5 - 2√3 

⇒ x² = (5 - 2√3)² 

⇒ x² = 5² + (2√3)² – 2(5) (2√3) 

⇒ x² = 25 + 12 – 20√3 

⇒ x² – 37 = – 20√3

⇒ \(\frac{37-x^2}{20}\) = √3

Since x is a rational number, x² is also a rational number. 

⇒ 37 - x² is a rational number 

⇒ \(\frac{37-x^2}{20}\) is a rational number 

⇒√3 is a rational number But √3 is an irrational number, which is a contradiction. 

Hence, our assumption is wrong. 

Thus, (5 - 2√3) is an irrational number.

Let us assume that (5 – 2√3) is a rational.

Subtract given number form 5, considering 5 is a rational number.

As we know, Difference of two rational numbers is a rational.

5 – (5 – 2√3) is rational

2√3 is rational

Which is only possible if 2 is rational and √3 is rational

As we know, product of two rational number is rational.

But the fact is √3 is an irrational.

Which is contradict to our assumption.

(5 – 2√3) is an irrational number.

To prove:

5 - 2√3 is an irrational number.

Solution:

Let assume that 5 - 2√3 is rational.

Therefore it can be expressed in the form of \(\frac{p}{q}\), where p and q are integers and q ≠ 0

Therefore we can write 5 - 2√3 = \(\frac{p}{q}\)

2√3 = 5 - \(\frac{p}{q}\) ⇒ √3 = \(\frac{5q-p}{2q}\)

 \(\frac{5q-p}{2q}\)is a rational number as p and q are integers.

This contradicts the fact that √3 is irrational, so our assumption is incorrect.

Therefore 5 - 2√3 is irrational.

Note:

Sometimes when something needs to be proved, prove it by contradiction.

Where you are asked to prove that a number is irrational prove it by assuming that it is rational number and then contradict it.

240 = 228 x 1 + 12 

and 228=12 x 19 + 0

Hence, HCF of 240  and 228 = 12

Given numbers are 4052 and 12576

Here, 12576 > 4052

So, we divide 12576 by 4052

By using Euclid’s division lemma, we get

12576 = 4052 × 3 + 420

Here, r = 420 ≠ 0.

On taking 4052 as dividend and 420 as the divisor and we apply Euclid’s division lemma, we get

4052 = 420 × 9 + 272

Here, r = 272 ≠ 0

On taking 420 as dividend and 272 as the divisor and again we apply Euclid’s division lemma, we get

420 = 272 × 1 + 148

Here, r = 148 ≠ 0

On taking 272 as dividend and 148 as the divisor and again we apply Euclid’s division lemma, we get

272 = 148 × 1 + 124

Here, r = 124 ≠ 0.

On taking 148 as dividend and 124 as the divisor and we apply Euclid’s division lemma, we get

148 = 124 × 1 + 24

Here, r = 24 ≠ 0

So, on taking 124 as dividend and 24 as the divisor and again we apply Euclid’s division lemma, we get

124 = 24 × 5 + 4

Here, r = 4 ≠ 0

So, on taking 24 as dividend and 4 as the divisor and again we apply

Euclid’s division lemma, we get

24 = 4 × 6 + 0

The remainder has now become 0, so our procedure stops. Since the divisor at this last stage is 4, the HCF of 4052 and 12576 is 4.

Correct option is (B) x + y

\(x=log_2\,3\) and \(y=log_2\,5\)

\(log_2\,15=log_2\,(3\times5)\)

\(=log_2\,3+log_2\,5\) = x+y  \((\because\) log AB = log A + log B)

Correct option is B) x + y

Correct option is (A) 2⁴ x 3²

H.C.F. of \(3^7\times5^3\times2^4\) and \(3^2\times7^4\times2^8\)

= Highest common factor of \(3^7\times5^3\times2^4\) and \(3^2\times7^4\times2^8\)

= \(3^2\times2^4\)

Correct option is A) 2⁴ × 3²

Correct option is (A) 1

\(\frac{1}{log_2\,log_2\,log_2\,16}\) \(=\frac{1}{log_2\,log_2\,log_2\,2^4}\)

\(=\frac{1}{log_2\,log_2\,4\,log_2\,^2}\)

\(=\frac{1}{log_2\,log_2\,2^2}\)   \((\because log_2\,^2=1)\)

\(=\frac{1}{log_2\,2\,log_2\,2}\)

\(=\frac{1}{log_2\,2}\)   \((\because log_2\,^2=1)\)

= 1

Correct option is (A) 1

Correct option is (C) log (9/8)

2 log 3 – 3 log 2 \(=log\,3^2-log\,2^3\)   \((\because log\,a^n=n\,log\,a)\)

= log 9 - log 8

= \(log\,\frac98\)    \((\because log\,\frac AB\) = log A - log B)

Correct option is C) log ( 9/8)

Correct option is (C) -1

\(log_2\,(log_{25}\,5)\) \(=log_2\,(\frac1{log_{5}\,25})\)   \((\because log_b\,a=\frac1{log_a\,b})\)

\(=log_2\,(\frac1{log_{5}\,5^2})\)

\(=log_2\,(\frac1{2\,log_{5}\,5})\)   \((\because log\,a^n=n\,log\,a)\)

\(=log_2\,(\frac1{2})\)    \((\because log_a\,a=1)\)

\(=log_2\,(2^{-1})\)

\(=-1\,log_2\,2\) = -1

Correct option is C) -1

Correct option is (B) 1.4771

log 15 + log 2 \(=log\,(15\times2)\)   \((\because\)  log A + log B = log AB)

= log 30

\(=log\,(3\times10)\)

= log 3 + log 10

= 0.4771 + 1   \((\because log\,10=log_{10}\,10=1)\)

= 1.4771

Correct option is B) 1.4771

Correct option is (A) 1.4771

log 15 + log 2 \(=log\,(15\times2)\)

= log 30

\(=log\,(10\times3)\)

= log 10 + log 3

= 1 + 0.4771   \((\because log_{10}\,10=1\,\&\,log_{10}\,3=0.4771)\)

= 1.4771

Correct option is (B) 1

\(\frac{1}{log_x\,xy} + \frac{1}{log_y\,xy} \) \(=\frac{1}{log_x\,x+log_x\,y} + \frac{1}{log_y\,x+log_y\,y} \)   \((\because\) log AB = log A + log B)

\(=\frac{1}{1+log_x\,y} + \frac{1}{1+log_y\,x} \)   \((\because log_aa=1)\)

\(=\frac{1}{1+log_x\,y} + \frac{1}{1+\frac1{log_x\,y}} \)    \((\because log_a\,b=\frac1{log_b\,a})\)

\(=\frac{1}{1+log_x\,y} + \frac{log_x\,y}{1+log_x\,y} \)

\(=\frac{1+log_x\,y}{1+log_x\,y}=1\)

Correct option is B) 1

Given: x² + y² = 25xy 

We know that (x + y)² = x² + y² + 2xy 

= 25xy + 2xy    [∵ x² + y² = 25xy given] 

(x + y)² = 27xy 

Taking ‘log’ on both sides 

log (x + y)² = log 27xy

2 log (x + y) = log 27 + log x + log y 

= log 3³ + log x + log y 

⇒ 2 log (x + y) = 3log3 + log x + log y

No.

Justification:

Consider the positive integer 3q + 1, where q is a natural number.

(3q + 1)2 = 9q² + 6q + 1

= 3(3q² + 2q) + 1

= 3m + 1, (where m is an integer which is equal to 3q2 + 2q.

Thus (3q + 1)² cannot be expressed in any other form apart from 3m + 1.

i.  \(\frac{3}{5}\sqrt{10}\) 

3/5√10

 = \(\frac{3}{5}\sqrt{10}\) x \(\sqrt{10}\)

 = \(\frac{3}{5}\times10\) 

= 3 x 2 

= 6, which is a rational number.

∴ √10 is the simplest form of the rationalising factor of \(\frac{3}{5}\sqrt{10}\) .

ii. \(3\sqrt{72}\) 

3√72

 = \(3\sqrt{36\times2}\) =3 x \(6\sqrt{2}\) = \(18\sqrt{2}\)

Now, 18√2 x √2 = 18 x 2 = 36, which is a rational number.

∴ √2 is the simplest form of the rationalising factor of \(3\sqrt{72}\) 

iii. \(4\sqrt{11}\) 

4√11 

\(4\sqrt{11}\) x \(\sqrt{11}\) = 4 x 11 =44 which is a rational number.

∴ √11 is the simplest form of the rationalising factor of \(4\sqrt{11}\).