Math, asked by kaushika73, 7 hours ago

please do its urgent (1,3,4)​

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Answers

Answered by tennetiraj86
0

Step-by-step explanation:

Solutions:-

1) Given that:

7 √(1/3) - 2 1/3 √(1/3) +3√147

=>7(1/√3) - (7/3)(1/√3) +3√(3×7×7)

=> (7/√3) - (7/3√3) +3×7√3

=> (7/√3) - (7/3√3) + 21√3

On Rationalising the denominators then

=>[ 7√3/(√3×√3)]-[7√3/(3×√3×√3) +21√3

The Rationalising factor of√3 =√3

=> (7√3/3)-[7√3/(3×3)]+21√3

=> (7√3/3)-(7√3/9)+21√3

LCM of 3 and 9 = 9

=> [(3×7√3)-(7√3)+(9×21√3)]/9

=> (21√3-7√3+189√3)/9

=>[(21-7+189)√3]/9

=> [(210-7)√3]/9

=> 203√3/9

3)

Given that :

x = 5-2√6

1/x = 1/(5-2√6)

The denominator = 5-2√6

The Rationalising factor of 5-2√6 = 5+2√6

On Rationalising the denominator then

=> 1/x = [1/(5-2√6)]×[(5+2√6)/5+2√6)]

=> 1/x = (5+2√6)/[5-2√6)(5+2√6)]

=> 1/x = (5+2√6)/[5²-(2√6)²]

Since (a+b)(a-b) = a²-b²

Where , a = 5 and b = 2√6

=>1/x = (5+2√6)/(25-24)

=> 1/x = (5+2√6)/1

=> 1/x = 5+2√6

Now,

x +(1/x)

=> 5-2√6+5+2√6

=> 5+5

=> 10

Therefore, x+(1/x) = 10

On squaring both sides then

=> [x+(1/x)]² = 10²

=> x²+2(x)(1/x)+(1/x)² =100

=> x²+2+(1/x)² = 100

=> x²+(1/x²) = 100-2

=> x²+(1/x²) = 98

Therefore, x²+(1/x²) = 98

4)Given number = √5

Let us assume that √5 is a rational number.

It must be in the form of p/q

Where, p and q are integers and q≠0

Let √5 = a/b (a,b are co primes)

On squaring both sides

=> (√5)² = (a/b)²

=> 5 = a²/b²

=> 5b² = a² ---------(1)

=> b² = a²/5

=> 5 divides a²

=>5 divides a also

=> 5 is a multiple of a ------(2)

Put a = 5c in (1) then

5b² = (5c)²

=> 5b² = 25c²

=> b² = 5c²

=> c² = b²/5

=> 5 divides b²

=> 5 divides also b

=> 5 is a multiple of b --------(3)

From (2)&(3)

We have,

5 is a multiple of both a and b

=> 5 is a common multiple of a and b

but a and b are co primes

They have only one common factor that is 1

This contradicts to our assumption that is√5 is a rational number.

So, √5 is not a rational number.

√5 is an irrational number.

Hence, Proved.

and

Given number = 3√5

Let us assume that 3√5 is a rational number.

It must be in the form of p/q

Where, p and q are integers and q≠0

Let 3√5 = a/b (a,b are co primes)

=> √5 = (a/b)/3

=> √5 = a/3b

=> √5 is in the form of p/q

=>√5 is a rational number

But √5 is not a rational number.

It is an irrational number

This contradicts to our assumption that is 3√5 is a rational number.

So, 3√5 is not a rational number.

3√5 is an irrational number.

Hence, Proved.

Note :-

The product of a rational and an irrational is an irrational number.

Answer:-

1) 203√3/9

3)x+(1/x) = 10 and x²+(1/x²) = 98

4) √5 and 3√5 are irrational numbers

Used formulae:-

  • The Rationalising factor of a-√b is a+√b

Used Method:-

  • Method of Contradiction ( Indirect method)
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