Question

plzz tell answer whose answer will correct I will make him brainlist​

Answer 1

Answer:

7.372/3.372 or 6+√3/2+√3

Answer 2

Since we can write

$1 \div x = {x}^{ - 1}$

therefire new equation will be

$x - {x}^{ - 1}$

substituting value of x

$(2 + \sqrt{3} ) - (2 + \sqrt{3} ) {}^{ - 1}$

$2 + \sqrt{3} - 1 \div 2 + \sqrt{3}$

$(2 + \sqrt{3} ) {}^{2} - 1 \div 2 + \sqrt{3}$

applying

$(a + b) {}^{2} = a {}^{2} + b {}^{2} + 2ab$

$2 {}^{2} + (\sqrt{3 } ) {}^{2} + 2(2) \times \sqrt{3} - 1 \div 2 + \sqrt{3}$

$4 + 3 + 4 \sqrt{3} - 1 \div 2 + \sqrt{3}$

$6 + 4 \sqrt{3} \div 2 + \sqrt{3}$

rationalizing the denominator

$6 + 4\sqrt{3} \div 2 + \sqrt{3} \: \: \: \: \: \: \: \: \: \: \times 2 - \sqrt{3} \div 2 - \sqrt{3}$

$(6 + 4 \sqrt{3} )(2 - \sqrt{3} ) \div (2) {}^{2} - ( \sqrt{3} ) {}^{2}$

since

$(x + y)(x - y) = x {}^{2} - y {}^{2}$

$6(2 - \sqrt{3} ) + 4 \sqrt{3} (2 - \sqrt{3} ) \div 4 - 3$

$12 - 6 \sqrt{3} + 8 \sqrt{3} - 12 \div 1$

$8 \sqrt{3} - 6 \sqrt{3}$

$2 \sqrt{3}$

$substituting \: value \: of \: \sqrt{3} = 1.732$

$2(1.732) = 3.464$

HOPE IT HELPS