Question

Rationalise
the denomination
4

कृष्णा लाइव द डायमीटर ऑफ 4 रूट 3 प्लस 5 रूट 2 बाय टू रूट 3 प्लस 3 रूट 2 ​

Answer 1

Answer:

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Answer 2

$\boxed{\dfrac{3+ \sqrt{6}}{3}}$

Step-by-step explanation:

To Rationalize:

$\dfrac{4 \sqrt{3} +5 \sqrt{2}}{2 \sqrt{3} + 3 \sqrt{2}}$

Solution:

$\dfrac{4 \sqrt{3} +5 \sqrt{2}}{2 \sqrt{3} + 3 \sqrt{2} } \\ \\ \frac{(4 \sqrt{3} + 5 \sqrt{2})(2 \sqrt{3} - 3 \sqrt{2})}{(2 \sqrt{3} + 3 \sqrt{2})(2 \sqrt{3} - 3 \sqrt{2})} \because {a}^{2} - {b}^{2} = (a + b)(a - b) \\ \\ \frac{4 \sqrt{3} \times 2 \sqrt{3} - 4 \sqrt{3} \times 3 \sqrt{2} + 5 \sqrt{2} \times 2 \sqrt{3} - 5 \sqrt{2} \times 3 \sqrt{2}}{ {(2 \sqrt{3})}^{2} - {(3 \sqrt{2}) }^{2} } \\ \\ \frac{8.3 - 15.2 - 12 \sqrt{6} + 10 \sqrt{6} }{4.3 - 9.2} \\ \\ \frac{24 - 30 - 2 \sqrt{6}}{12 - 18} \\ \\ \frac{ - 6 - 2 \sqrt{6} }{ - 6} \\ \\ \frac{ - 2( 3 + \sqrt{6})}{ - 6} \\ \\ \frac{ 3+ \sqrt{6}}{3}$

Therefore, the required answer is (3+√6)/3.

Extra Information:

Rationalizing the denominator is a process by which we can write the irrational denominator in the form of Rational no.

For Rationalizing we use a Conjugate surds or Rationalizing factor which is a factor of the irrational denominator.

e.g conjugate surd of √a is √a and the conjugate surds √a - b is √a + b.