Question

plzz help me oue with this question

Answer 1

Hello Mate!

$x = \frac{1}{3 - 2 \sqrt{2} } \times \frac{3 + 2 \sqrt{2} }{3 + 2 \sqrt{2} } \\ \frac{3 + 2 \sqrt{2} }{9 - 8} = 3 + 2 \sqrt{2}$

$y = \frac{1}{3 + 2 \sqrt{2} } \times \frac{3 - 2 \sqrt{2} }{3 - 2 \sqrt{2 } } \\ \frac{3 - 2 \sqrt{2} }{1} = 3 - 2 \sqrt{2}$
${y}^{2} = {(3 - 2 \sqrt{2} )}^{2} \\ = 9 + 8 - 12 \sqrt{2} \\ = 17 - 12 \sqrt{2} \\ x {y}^{2} = (3 + 2 \sqrt{2} )(17 - 12 \sqrt{2} ) \\ 51 - 36 \sqrt{2} + 34 \sqrt{2} - 48 \\ 3 - 2 \sqrt{2}$
${x}^{2} = {(3 + 2 \sqrt{2} )}^{2} \\ = 9 + 8 + 12 \sqrt{2} \\ = 17 + 12 \sqrt{2} \\ {x}^{2} y = (17 + 12 \sqrt{2} )(3 - 2 \sqrt{2} ) \\ 51 - 34 \sqrt{2} + 36 \sqrt{2} - 48 \\ 3 + 2 \sqrt{2}$
$x {y}^{2} + {x}^{2} y \\ = 3 - 2 \sqrt{2} + 3 + 2 \sqrt{2} \\ = 6$
Hope it helps☺!

Answer 2

The answer is given below :

Given that,

x = 1/(3 - 2√2) and y = 1/(3 + 2√2)

We rationalise the denominators of the values of x and y by multiplying both the numerators and denominators by the conjugate irrational numbers (3 + 2√2) and (3 - 2√2) respectively.

So,

x = (3 + 2√2)/{(3 - 2√2)(3 + 2√2)}

= (3 + 2√2)/(9 - 8)

= (3 + 2√2)

and

y = (3 - 2√2)/{(3 + 2√2)(3 - 2√2)}

= (3 - 2√2)/(9- 8)

= (3 - 2√2)

Now,

xy

= (3 + 2√2)(3 - 2√2)

= 9 - 8

= 1

So,

xy² + x²y

= xy(y + x)

= 1 × (3 - 2√2 + 3 + 2√2)

= 1 × 6

= 6 [Answer]

IDENTITY RULE USED :

a² - b² = (a + b)(a - b)

Thank you for your question.