Question

If x,y x , y are real numbers such that (x+1+x2)(y+1+y2)=1 \big( x+\sqrt{1+x^2}\big) \big( y+\sqrt{1+y^2} \big) =1 , find the value of (x+y)2 (x+y)^2

Collected in the board: Calculate algebraic expressions

Steven Zheng posted 1 year ago

Answer

Let

n=x+1+x2 n = x+\sqrt{1+x^2}

The asking expression is simplified as

n(y+1+y2)=1 n(y+\sqrt{1+y^2})=1

y+1+y2=1n y+\sqrt{1+y^2} = \dfrac{1}{n}

1+y2=1ny \sqrt{1+y^2} = \dfrac{1}{n}-y

(1+y2)2=(1ny)2 \Big( \sqrt{1+y^2}\Big) ^2 = \Big( \dfrac{1}{n}-y\Big) ^2

1+y2=1n22yn+y2 1+y^2 = \dfrac{1}{n^2}- \dfrac{2y}{n}+y^2

1n22yn=1 \dfrac{1}{n^2}- \dfrac{2y}{n}=1

2y=n1n -2y = n-\dfrac{1}{n}

y=12(1nn) y = \dfrac{1}{2}(\dfrac{1}{n}-n)
(1)

1nn \dfrac{1}{n}-n

=1x+1+x2(x+1+x2) =\dfrac{1}{ x+\sqrt{1+x^2}}-( x+\sqrt{1+x^2})

=x1+x2(x+1+x2)(x1+x2)(x+1+x2) =\dfrac{ x-\sqrt{1+x^2}}{( x+\sqrt{1+x^2})( x-\sqrt{1+x^2})} -( x+\sqrt{1+x^2})

=x1+x21(x+1+x2) =\dfrac{ x-\sqrt{1+x^2}}{-1} -( x+\sqrt{1+x^2})

=2x =-2x

y=x \therefore y = -x

(x+y)2=0 (x+y)^2=0

Steven Zheng posted 1 year ago

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