plzz help me oue with this question
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Answered by
3
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} 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}](https://tex.z-dn.net/?f=x+%3D++%5Cfrac%7B1%7D%7B3+-+2+%5Csqrt%7B2%7D+%7D++%5Ctimes++%5Cfrac%7B3+%2B+2+%5Csqrt%7B2%7D+%7D%7B3+%2B+2+%5Csqrt%7B2%7D+%7D++%5C%5C++%5Cfrac%7B3+%2B+2+%5Csqrt%7B2%7D+%7D%7B9+-+8%7D++%3D+3+%2B+2+%5Csqrt%7B2%7D+)
![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 = \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}](https://tex.z-dn.net/?f=y+%3D++%5Cfrac%7B1%7D%7B3+%2B+2+%5Csqrt%7B2%7D+%7D++++%5Ctimes++%5Cfrac%7B3+-+2+%5Csqrt%7B2%7D+%7D%7B3+-+2+%5Csqrt%7B2+%7D+%7D++%5C%5C++%5Cfrac%7B3+-+2+%5Csqrt%7B2%7D+%7D%7B1%7D++%3D+3+-+2+%5Csqrt%7B2%7D+)
![{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} {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}](https://tex.z-dn.net/?f=+%7By%7D%5E%7B2%7D++%3D++%7B%283+-+2+%5Csqrt%7B2%7D+%29%7D%5E%7B2%7D++%5C%5C++%3D+9+%2B+8+-+12+%5Csqrt%7B2%7D++%5C%5C++%3D+17+-+12+%5Csqrt%7B2%7D++%5C%5C+x+%7By%7D%5E%7B2%7D++%3D+%283+%2B+2+%5Csqrt%7B2%7D+%29%2817+-+12+%5Csqrt%7B2%7D+%29+%5C%5C+51+-+36+%5Csqrt%7B2%7D++%2B+34+%5Csqrt%7B2%7D++-+48+%5C%5C+3+-+2+%5Csqrt%7B2%7D+)
![{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}^{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}](https://tex.z-dn.net/?f=+%7Bx%7D%5E%7B2%7D++%3D++%7B%283+%2B+2+%5Csqrt%7B2%7D+%29%7D%5E%7B2%7D++%5C%5C++%3D+9+%2B+8+%2B+12+%5Csqrt%7B2%7D++%5C%5C++%3D+17+%2B+12+%5Csqrt%7B2%7D++%5C%5C++%7Bx%7D%5E%7B2%7D+y+%3D+%2817+%2B+12+%5Csqrt%7B2%7D+%29%283+-+2+%5Csqrt%7B2%7D+%29+%5C%5C+51+-+34+%5Csqrt%7B2%7D++%2B+36+%5Csqrt%7B2%7D++-+48+%5C%5C+3+%2B+2+%5Csqrt%7B2%7D+)
![x {y}^{2} + {x}^{2} y \\ = 3 - 2 \sqrt{2} + 3 + 2 \sqrt{2} \\ = 6 x {y}^{2} + {x}^{2} y \\ = 3 - 2 \sqrt{2} + 3 + 2 \sqrt{2} \\ = 6](https://tex.z-dn.net/?f=x+%7By%7D%5E%7B2%7D++%2B++%7Bx%7D%5E%7B2%7D+y+%5C%5C++%3D+3+-+2+%5Csqrt%7B2%7D++%2B+3+%2B+2+%5Csqrt%7B2%7D++%5C%5C++%3D+6)
Hope it helps☺!
Hope it helps☺!
Swarup1998:
Woah! Nice answer, dear! (:
Answered by
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.
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.
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