Question

please do it i have to submit my work cant understand

Answer 1

Answer:

Step-by-step explanation:

Answer 2

[tex]\mathfrak{\huge{Answer:-}} \\ \bold{ \frac{3 + 2 \sqrt{2} }{2 - \sqrt{2} } - \frac{3 - 2 \sqrt{2} }{2 + \sqrt{2} } = a + b \sqrt{2} } \\ \frac{3 + 2 \sqrt{2} }{2 - \sqrt{2} } \times \frac{2 + \sqrt{2} }{2 + \sqrt{2} } - \frac{3 - 2 \sqrt{2} }{2 + \sqrt{2} } \times \frac{2 - \sqrt{2} }{2 - \sqrt{2} }= a + b \sqrt{2} \\ \frac{(3 + 2 \sqrt{2})(2 + \sqrt{2})}{ (2 - \sqrt{2})(2 + \sqrt{2})} - \frac{(3 - 2 \sqrt{2})(2 - \sqrt{2}) }{(2 + \sqrt{2})(2 - \sqrt{2}) }= a + b \sqrt{2} \\ \frac{2(3 + 2 \sqrt{2}) + \sqrt{2} (3 + 2 \sqrt{2}) }{ {(2)}^{2} - {( \sqrt{2}) }^{2} } - \frac{2(3 - 2 \sqrt{2} ) - \sqrt{2} (3 - 2 \sqrt{2}) }{ {(2)}^{2} - {( \sqrt{2}) }^{2} }= a + b \sqrt{2} \\ \frac{6 + 4 \sqrt{2} + 3 \sqrt{2} + 4 }{4 - 2} - \frac{6 - 4 \sqrt{2} - 3 \sqrt{2} + 4 }{4 - 2} = a + b \sqrt{2} \\ \frac{10 + 7 \sqrt{2} }{2} - \frac{10 - 7 \sqrt{2} }{2} = a + b \sqrt{2} \\ \frac{(10 + 7 \sqrt{2} ) - (10 - 7 \sqrt{2} )}{2} = a + b \sqrt{2} \\ \frac{10 + 7 \sqrt{2} - 10 + 7 \sqrt{2} }{2} \\ \frac{14 \sqrt{2} }{2} = a + b \sqrt{2} \\ 7 \sqrt{2} = a + b \sqrt{2} \\ 0+ 7 \sqrt{2} = a + b \sqrt{2} \\ a = 0, \: b = 7 [/tex]