Question

Hey Mate, I am strucked in this problem between. Hopes that you can solve it with explanation.
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(thank you) ​

Answer 1

Answer:

- 2

Step-by-step explanation:

Given :

$\sf \dfrac{ - 3}{ \sqrt{3} + \sqrt{2} } - \dfrac{3 \sqrt{2} }{ \sqrt{6} + \sqrt{3} } + \dfrac{4 \sqrt{3} }{ \sqrt{6} + \sqrt{2} } = a \sqrt{2} + b \sqrt{3}$

Rationalising the denominator of each term of LHS

$\Rightarrow \sf \dfrac{ - 3( \sqrt{3} - \sqrt{2} )}{ (\sqrt{3} + \sqrt{2})( \sqrt{3} - \sqrt{2} )} - \dfrac{3 \sqrt{2}( \sqrt{6} - \sqrt{3} )}{ (\sqrt{6} + \sqrt{3})( \sqrt{6} - \sqrt{3} )} + \dfrac{4 \sqrt{3}( \sqrt{6} - \sqrt{2} ) }{( \sqrt{6} + \sqrt{2} )( \sqrt{6} - \sqrt{2}) } = a \sqrt{2} + b \sqrt{3}$

Using identity (x + y)(x - y) = x² - y² in denominators

$\Rightarrow \sf \dfrac{ - 3 \sqrt{3} + 3 \sqrt{2} }{ (\sqrt{3} ) ^{2} - ( \sqrt{2})^{2} } - \dfrac{3 \sqrt{12} - 3\sqrt{6} }{ (\sqrt{6} )^{2} - (\sqrt{3}) ^{2} } + \dfrac{4 \sqrt{18} - 4\sqrt{6} }{( \sqrt{6})^{2} - (\sqrt{2})^{2} } = a \sqrt{2} + b \sqrt{3}$

$\Rightarrow \sf \dfrac{ - 3 \sqrt{3} + 3 \sqrt{2} }{3 - 2} - \dfrac{3 \sqrt{12} - 3\sqrt{6} }{6 - 3 } + \dfrac{4 \sqrt{18} - 4\sqrt{6} }{6 - 2} = a \sqrt{2} + b \sqrt{3}$

$\Rightarrow \sf - 3 \sqrt{3} + 3 \sqrt{2} - \dfrac{3( \sqrt{12} - \sqrt{6} )}{3 } + \dfrac{4( \sqrt{18} - \sqrt{6} ) }{4} = a \sqrt{2} + b \sqrt{3}$

⇒ - 3√3 + 3√2 - ( √12 - √6 ) + √18 - √6 = a√2 + b√3

⇒ - 3√2 - √12 + √6 + √18 - √6 = a√2 + b√3

⇒ - 3√2 - √( 4 × 3 ) + √( 9 × 2 ) = a√2 + b√3

Using √xy = √x × √y

⇒ - 3√2 - ( √4 × √3 ) + ( √9 × √2 ) = a√2 + b√3

⇒ - 3√2 - 2√3 + 3√2 = a√2 + b√3

⇒ 0√2 + - 2√3 = a√2 + b√3

Comparing on both sides

• a = 0
• b = - 2

a + b = 0 + ( - 2 ) = 0 - 2 = - 2

Therefore the value of a + b is - 2.