Question

If
$(x + \sqrt{1 + x^{2} })(y + \sqrt{1 + {x}^{2} } ) = 1$
find,
${(x + y)}^{2020}$
​

Answer 1

Answer:

Ok i will be the first time m

Step-by-step explanation:

X7 uw 62 + +73 82 + 28

Answer 2

$\rm{\left(x+\sqrt{1+x^2}\right)\left(y+\sqrt{1+x^2}\right)=1}$

$\longrightarrow\rm{x\left(y+\sqrt{1+x^2}\right)+\sqrt{1+x^2}\left(y+\sqrt{1+x^2}\right)=1}$

$\longrightarrow\rm{xy+\sqrt{1+x^2}\:x+\sqrt{1+x^2}\:y+1+x^2=1}$

$\longrightarrow\rm{xy+\sqrt{1+x^2}\:x+\sqrt{1+x^2}\:y+1+x^2-1=1-1}$

$\longrightarrow\rm{x^2+\sqrt{1+x^2}\:x+xy+\sqrt{1+x^2}\:y=0}$

$\longrightarrow\rm{x\left(x+\sqrt{1+x^2}\right)+y\left(x+\sqrt{1+x^2}\right)=0}$

$\longrightarrow\rm{\left(\sqrt{1+x^2}+x\right)\left(x+y\right)=0}$

$\longrightarrow\rm{\sqrt{1+x^2}+x=0\quad \mathrm{or}\quad \:x+y=0}$

$\longrightarrow\rm{\sqrt{1+x^2}=-x\quad \mathrm{or}\quad \:x+y-y=0-y}$

$\longrightarrow\rm{1+x^2=x^2\quad \mathrm{or}\quad \:x=0-y}$

$\longrightarrow\rm{\mathrm{No\:Solution\:for}\:x\in \mathbb{R}\quad \mathrm{or}\quad \:x=-y\quad...\textrm{ So x is equal to -y}}$

$\rule{315}{2}$

$\rm{(x+y)^{2020}=(-y+y)^{2020}\quad...\:Since\:x=-y}$

$\longrightarrow\rm{(x+y)^{2020}=0^{2020}=0}$