Question

if x=3+2√2 and xy=1, then (x²+3xy+y²)/(x²-3xy+y²)

Answer 1

x = (3+ 2√2) and xy = 1

y = 1/x =1/( 3 + 2√2)
=( 3 - 2√2)/(3 + 2√2)(3 -2√2)
=(3 -2√2)/(3² - 2√2²)
=(3 -2√2)/(9 -8)
= 3 -2√2
hence ,
(x + y) = 3 +2√2 + 3 -2√2 = 6

now,

(x² + 3xy + y²) = {(x² + y²) + 3xy }

= { (x + y)² -2xy + 3xy }

= { (x + y)² + xy }

= { ( 6)² + 1}

= 37

similarly ,
(x² -3xy + y²)

= { (x + y)² -2xy -3xy }

={ (x + y)² -5xy }

= { (6)² -5×1 }

= { 36 -5}

= 31

hence ,
(x² +3xy + y²)/(x² -3xy + y²) = 37/31

Answer 2

x = 3+2√2

xy = 1

y = 1/x

→ = 1/3+2√2

→ = 1(3-2√2)/(3+2√2)(3-2√2)

→ = 3-2√2/{3²-(2√2)²}

→ = 3-2√2/(9-4(2))

→ = 3-2√2/9-8

→ y = 3-2√2

x+y = 3+2√2+3-2√2

= 6

Let's solve first,

(x²+3xy+y²)

→ (x²+y²+3xy)

→ {(x+y)²-2xy}+3xy

→ {6²-2(1)} + 3(1)

→ 36-2+3

→ 37

Next,

(x²-3xy+y²)

→ (x²+y²-3xy)

→ {(x+y)²-2xy}-3xy

→ 6²-2(1)-3(1)

→ 36-2-3

→ 31

(x²+3xy+y²)/(x²-3xy+y²) = 37/31

Hope it helps …