Question

rationalise the denominator
pls
it's urgent​

Answer 1

Answer:

answer is 0

Step-by-step explanation:

we need to rationalise all the three terms separately step by step to reach the final answer.

Answer 2

Answer:

0.

Step-by-step explanation:

Given:-

[3√2/(√3+√6)] - [4√3/(√6+√2)] + [√6/(√2+√3)]

To find out:-

Rationalise all the denominator.

Solution:-

We have,

(3√2/(√3+√6)] - [4√3/(√6+√2)] + [√6/(√2+√3)]

Here we will see that in expression given three terms that is First term , Sec term and Third term. So we can solve one by one and then rationalising all the three terms denominator. We arrange all the rationalised values according to the given question and simplify that and last you will get your final answer.

Let's solve!

First term:

(3√2)/(√3+√6)

→ (3√2)/√3(1+√2)

→ √6/(√2+1)

The denomination = √2 +1

We know that

Rationalising factor of √a + b = √a - b

So,rationalising factor of √2+1 = √2-1

On rationalising the denominator them

→ [√6/(√2+1)] × [(√2-1)/(√2-1)

→ [√6(√2-1)]/[(√2+1)(√2-1)]

Now, applying algebraic Identity in denominator because it is in the form of;

(a + b)(a-b) = a^2 - b^2

Where we have to put in our expression a = √2 and b = 1, we get

→ [√6(√2-1)]/[(√2)^2 - (1)^2]

→ [√6(√2-1)]/(2 - 1)

→ [(√6(√2-1)]/1

→ 2√3 - √6

Second term:

(4√3)/(√6+√2)

→ (4√3)/√2(√3+1)

→ (2√6)/(√3+1)

The denomination = √3+1

We know that

Rationalising factor of √a + b = √a - b

So,rationalising factor of √3+1 = √3-1

On rationalising the denominator them

→ [(2√6)/(√3+1)] × [(√3-1)/(√3-1)]

→ [(2√6)(√3-1)]/[(√3+1)(√3-1)]

Now, applying algebraic Identity in denominator because it is in the form of;

(a + b)(a-b) = a^2 - b^2

Where we have to put in our expression a = √3 and b = 1 , we get

→ [(2√6)(√3-1)]/[(√3)^2 - (1)^2]

→ [(2√6)(√3-1)]/(3 - 1)

→ [(2√6)(√3-1)]/ 2

→ 2(3√2 - √6)/2

→ 3√2 - √6

Third term:

√6 /(√2 + √3)

→ √6/(√3+√2)

The denomination = √3 + √2

We know that

Rationalising factor of √a + √b = √a - √b

So,rationalising factor of √3+√2 = √3-√2

On rationalising the denominator them

→ [√6/(√3+√2] × [(√3-√2)/(√3-√2)]

→ [√6(√3-√2)]/[(√3+√2)(√3-√2)]

Now, applying algebraic Identity in denominator because it is in the form of;

(a + b)(a-b) = a^2 - b^2

Where we have to put in our expression a=√3 and b = √2 , we get

→ [√6(√3-√2)]/[(√3)^2 - (√2)^2]

→ [√6(√3-√2)]/(3-1)

→ √6(√3-√2)/1

→ 3√3 - 2√3

Now arranging all the rationalised values according to the given question and simplify;

Hence, [3√2/(√3+√6)] - [4√3/(√6+√2)] + [√6/(√2+√3)]

Putting value in their places, we get

= (2√3-√6) - (3√2-√6) + (3√2-2√3)

Now, opening brackets, we get

= 2√3 - √6 - 3√2 + √6 + 3√2 - 2√3

= 0. Ans.

L.H.S = R.H.S