Question

√2-1÷√2-1=x+y√2. plz give me answer very fast​

Answer 1

Step-by-step explanation:

Answer:

x^2+y^2+xy=35x

2

+y

2

+xy=35

Step-by-step explanation:

Given : x=\frac{\sqrt2+1}{\sqrt2-1}x=

2

−1

2

+1

and y=\frac{\sqrt2-1}{\sqrt2+1}y=

2

+1

2

−1

To find : The value of x^2+y^2+xyx

2

+y

2

+xy

Solution :

First we solve the value of x and y be rationalizing,

The value of x,

x=\frac{\sqrt2+1}{\sqrt2-1}x=

2

−1

2

+1

x=\frac{\sqrt2+1}{\sqrt2-1}\times\frac{\sqrt2+1}{\sqrt2+1}x=

2

−1

2

+1

×

2

+1

2

+1

x=\frac{(\sqrt2+1)^2}{(\sqrt2)^2-1^2}x=

(

2

)

2

−1

2

(

2

+1)

2

x=\frac{2+1+2\sqrt2}{2-1}x=

2−1

2+1+2

2

x=3+2\sqrt2x=3+2

2

The value of y,

y=\frac{\sqrt2-1}{\sqrt2+1}y=

2

+1

2

−1

y=\frac{\sqrt2-1}{\sqrt2+1}\times\frac{\sqrt2-1}{\sqrt2-1}y=

2

+1

2

−1

×

2

−1

2

−1

y=\frac{(\sqrt2-1)^2}{(\sqrt2)^2-1^2}y=

(

2

)

2

−1

2

(

2

−1)

2

y=\frac{2+1-2\sqrt2}{2-1}y=

2−1

2+1−2

2

y=3-2\sqrt2y=3−2

2

Now, Substitute the value of x and y in the expression

x^2+y^2+xyx

2

+y

2

+xy

=(3+2\sqrt2)^2+(3-2\sqrt2)^2+(3+2\sqrt2)(3-2\sqrt2)=(3+2

2

)

2

+(3−2

2

)

2

+(3+2

2

)(3−2

2

)

=9+8+12\sqrt2+9+8-12\sqrt2+3^2-(2\sqrt2)^2=9+8+12

2

+9+8−12

2

+3

2

−(2

2

)

2

=17+17+9-8=17+17+9−8

=35=35

Therefore, x^2+y^2+xy=35x

2

+y

2

+xy=35