Question

can y'all please solve these ​

Answer 1

Answer:

$1) {16}^{ \frac{3}{4} } \div {16}^{ \frac{1}{4} } = {16}^{ \frac{3}{4} - \frac{1}{4} } = {16}^{ \frac{3 - 1}{4} } = {16}^{ \frac{2}{4} } = {16}^{ \frac{1}{2} } = \sqrt{16} = 4 \\ 2) \frac{ \sqrt{2} + 1 }{ \sqrt{2} - 1 } = a + b \sqrt{2} \\ \frac{ \sqrt{2} + 1}{ \sqrt{2} - 1 } \times \frac{ \sqrt{2} + 1 }{ \sqrt{2} + 1 } = a + b \sqrt{2} \\ \frac{ {( \sqrt{2} + 1) }^{2} }{ {( \sqrt{2} )}^{2} - {(1)}^{2} } = a + b \sqrt{2} \\ \frac{2 + 1 + 2 \sqrt{2} }{2 - 1} = a + b \sqrt{2} \\ \frac{3 + 2 \sqrt{2} }{1} = a + b \sqrt{2} \\ 3 + 2 \sqrt{2} = a + b \sqrt{2} \\ a = 3 \: and \: b = 2 \\ 3)x = \sqrt{5} + 2 \\ x - \frac{1}{x} \\ = \sqrt{5} + 2 - \frac{1}{ \sqrt{5} + 2 } \\ = \frac{ {( \sqrt{5} + 2) }^{2} - 1}{ \sqrt{5} + 2} \\ = \frac{5 + 4 + 4 \sqrt{5} }{ \sqrt{5} + 2 } \\ = \frac{9 + 4 \sqrt{5} }{ \sqrt{5} + 2 } \\ 4) \frac{2 + \sqrt{3} }{2 - \sqrt{3} } - \frac{2 - \sqrt{3} }{2 + \sqrt{3} } \\ = \frac{2 + \sqrt{3} }{2 - \sqrt{3} } \times \frac{2 + \sqrt{3} }{2 + \sqrt{3} } - \frac{2 - \sqrt{3} }{2 + \sqrt{3} } \times \frac{2 - \sqrt{3} }{2 - \sqrt{3} } \\ = \frac{ {(2 + \sqrt{3} )}^{2} }{ {(2)}^{2} - {( \sqrt{3} )}^{2} } - \frac{ {(2 - \sqrt{3} )}^{2} }{ {(2)}^{2} - {( \sqrt{3} )}^{2} } \\ = \frac{4 + 3 + 4 \sqrt{3} }{4 - 3} - \frac{4 + 3 - 4 \sqrt{3} }{4 - 3} \\ = \frac{7 + 4 \sqrt{3} }{1} - \frac{(7 - 4 \sqrt{3} )}{1} \\ = 7 + 4 \sqrt{3} - 7 + 4 \sqrt{3} \\ = 7 - 7 + 4 \sqrt{3} + 4 \sqrt{3} \\ = 8 \sqrt{3}$