Question

answer all of 2nd

all four​

Answer 1

Answer:

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Answer 2

• 2nd

Rationalising the denominator

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

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

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

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

$\sqrt{2} (3 \sqrt{2} + 2 \sqrt{3} ) + \sqrt{3} (3 \sqrt{2} + 2 \sqrt{3} ) \\ \div \\ 9(2) - 4(3)$

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

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

$12 + 5 \sqrt{6} \\ \div \\ 6$

$12 \div 6 + 5 \sqrt{6} \div 6 = a - b \sqrt{6}$

$2 + 5 \sqrt{6} = a - b \sqrt{6}$

comparing both sides we get

$a = 2 \\ b = - 5 \div 6 = 0.833$

• 4th

rationalise both terms on left hand side

$(7 + \sqrt{5} \div 7 - \sqrt{5} \times 7 + \sqrt{5} \div 7 + \sqrt{5} )$

$(7 + \sqrt{5} ) {}^{2} \\ since \\ (x + y)(x + y) = (x + y) {}^{2} \\ also \\ (7 - \sqrt{5} )(7 + \sqrt{5})$

$49 + 5 + 14 \sqrt{5} \div 7 {}^{2} - \sqrt{5 {}^{2} } \\ since \\ (x + y) {}^{2} = x {}^{2} + y {}^{2} + 2xy \\ (x + y)(x - y) = x {}^{2} - y {}^{2}$

$54 + 14 \sqrt{5} \\ \div \\ 49 - 5$

$54 + 14 \sqrt{5} \\ \div \\ 44$

$54 \div 44 + 14 \sqrt{5} \div 44$

Solving 2nd term

$- (7 - \sqrt{5} \div 7 + \sqrt{5} )$

Taking only bracket

$7 - \sqrt{5} \times 7 - \sqrt{5} \\ \div \\ 7 + \sqrt{5} \times 7 - \sqrt{5}$

$(7 - \sqrt{5} ) {}^{2} \\ \div \\ 7 {}^{2} - {(\sqrt{5} ) }^{2}$

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

$49 + 5 - 14 \sqrt{5} \\ \div \\ 49 - 5$

$54 - 14 \sqrt{5} \\ \div \\ 44$

The new euation will be as follows

$54 + 14 \sqrt{5} - (54 - 14 \sqrt{5} ) \\ \div 44 \: \: \: \: \: \: \div (44)$

$54 + 14 \sqrt{5} - (54 - 14 \sqrt{5} ) \\ \div \\ 44$

$54 + 14 \sqrt{5} - 54 + 14 \sqrt{5} \\ \div \\ 44$

$2(14 \sqrt{5} ) \div 44$

$28 \sqrt{5} \\ \div \\ 44$

$\frac{7 \sqrt{5} }{11}$

this equation was equal to

$a + \frac{7b \sqrt{5} }{11}$

comparing we get

a=0

$b = 1$

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