Question

Rationalise the denominator of each of the following 2√3-√5/2√2+3√3

Answer 1

Answer:

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Answer 2

Step-by-step explanation:

Given that:-

$\frac{2 \sqrt{3} - \sqrt{5} }{2 \sqrt{2} + 3 \sqrt{3} }$

What to do:-

To rationalise the denominator.

Solution:-

We have,

$\frac{2 \sqrt{3} - \sqrt{5} }{2 \sqrt{2} + 3 \sqrt{3} }$

The denominator is 2√2+3√3. Multiplying the numerator and denomination by 2√2-3√3,

We get,

$⟹ \frac{2 \sqrt{3} - \sqrt{5} }{2 \sqrt{2} + 3 \sqrt{3} } \times \frac{2 \sqrt{2} - 3 \sqrt{3} }{2 \sqrt{2} - 3 \sqrt{3} }$

$⟹ \frac{(2 \sqrt{3} - \sqrt{5})(2 \sqrt{2} - 3 \sqrt{3} )}{(2 \sqrt{2} + 3 \sqrt{3})(2 \sqrt{2} - 3 \sqrt{3}) }$

⬤ Applying Algebraic Identity

(a+b)(a-b) = a² - b² to the denominator

We get,

$⟹ \frac{(2 \sqrt{2} \times 2 \sqrt{2}) + ( 2 \sqrt{3} \times - 3 \sqrt{3} ) - ( \sqrt{5} \times 2 \sqrt{2} ) + ( \sqrt{5} \times 3 \sqrt{3} ) }{(2 \sqrt{2} {)}^{2}(3 \sqrt{3} {)}^{2} }$

$⟹\frac{4 \sqrt{6} - 6 \times 3 - 2 \sqrt{10} + 3 \sqrt{15} }{4 \times 2 - 9\times 3}$

$⟹ \frac{4 \sqrt{6} - 18 - 2 \sqrt{10} + 3 \sqrt{15} }{8 - 27}$

$⟹ \frac{ - (18 - 3 \sqrt{15} + 2 \sqrt{10} - 4 \sqrt{6} ) }{ - 19}$

$⟹ \frac{ \cancel \red{- }18 - 3 \sqrt{15} + 2 \sqrt{10} - 4 \sqrt{6} } { \cancel\red{- }19}$

$⟹ \frac{18 - 3 \sqrt{15} + 2 \sqrt{10} - 4 \sqrt{6} ) }{ 19} \: \: \tt \red{Ans. }$

Hence, the denominator is rationalised.

Know more Algebraic Identities:-

(a+ b)² = a² + b² + 2ab

( a - b )² = a² + b² - 2ab

( a + b )² + ( a - b)² = 2a² + 2b²

( a + b )² - ( a - b)² = 4ab

( a + b + c )² = a² + b² + c² + 2ab + 2bc + 2ca

a² + b² = ( a + b)² - 2ab

(a + b )³ = a³ + b³ + 3ab ( a + b)

( a - b)³ = a³ - b³ - 3ab ( a - b)

If a + b + c = 0 then a³ + b³ + c³ = 3abc

I hope it's help you..☺