Question

if x=√2+1/√2-1 and y=√2-1/√2+1 find the value of x^2+y^2+xy​

Answer 1

Step-by-step explanation:

=> x = √2 + 1/ √2 +1

= (√2 + 1) (√2+1)/ (√2 -1) (√2+1)

= (√2+1)^2 / {(√2)^2 - (1)^2}

=(2 + 1 +2√2( / (2-1)

= 3 + 2 √2

=> y = √2-1 / √2 +1

= 3 - 2 √2 (by applying same process above for x )

=> xy = (√2 +1 / √ 2-1) × ( √2-1) / (√2+1) = 1

=> x+ y = 3 + 2 √2 + 3 -2√2 = 6

======> x^2 + y^2 + xy

= x^2 + y^2 + xy + xy - xy

= x^2 + 2xy + y^2 - xy

= (x+y)^2 -xy

= 6^2 -1

= 36 -1

= 35

Answer 2

Step-by-step explanation:

x^2+y^2+xy

(-/2+1/-/2-1)^2+(-/2-1/-/2+1)^2+(-/2+1/-/2-1)×(-/2+1/-/2+1)

2+1/2+1+2+1/2+1+2+1-1

8+1/2+1/2

8+2/2

16/2+2/2

18/2

9