Math, asked by saresmandi, 2 months ago

plz solve this problem ​

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Answers

Answered by simarjeetkaursidhu41
2

Answer:

weak in math don't know the ans

Answered by MrSovereign
3

Given:-

  • \bold{xyz = 1}

Required To Prove:-

  • \bold{\frac{1}{1+x+\frac{1}{y}}+\frac{1}{1+y+\frac{1}{z}}+\frac{1}{1+z+\frac{1}{x}} = 1}

Proof:-

⌘\;{\pink{\sf{\frac{1}{a} = (a)^{-1}}}}

→\;\bold{ \frac{1}{1 + x +  \frac{1}{y} }  +  \frac{1}{1 + y +  \frac{1}{z} }  +  \frac{1}{1 + z +  \frac{1}{x} } }

\bold{→\;( {1 + x +  {y}^{ - 1}) }^{ - 1}  +  {(1 + y + {z}^{ - 1} ) }^{ - 1}  +  ({1 + z +  {x}^{ - 1} )}^{ - 1}}

WKT

xyz = 1

Then,

➝ xz = \sf{ \frac{1}{y} }

➝ xz = \sf{ y^{-1}}

Similarly

  • zy = x^{-1}
  • xy = y^{-1}

\bold{→\; {(1 + x + xz)}^{ - 1}  +  {(1 + y + yx)}^{ - 1}  +  {(1 + z + zy)}^{ - 1} }

\bold{→\; {(1 + x + xz)}^{ - 1}  +  {(xyz + y + yx)}^{ - 1}  +  {(1 + z + zy)}^{ - 1} }

\bold{→\; {(1 + x + xz)}^{ - 1}  +  {y^{-1}(1 + y + zx)}^{ - 1}  +  {(1 + z + zy)}^{ - 1} }

\bold{→\; {(1 + x + xz)}^{ - 1}  (1+y^{-1}) +  {(1 + z + zy)}^{ - 1} }

\bold{→\; {(xyz + x + xz)}^{ - 1}  (1+y^{-1}) +  {(1 + z + zy)}^{ - 1} }

\bold{→\; {x^{-1}(1+z+yz)}^{ - 1}  (1+y^{-1}) +  {(1 + z + zy)}^{ - 1} }

\bold{→\;(1+z+yz)^{-1} [x^{-1}+(xy)^{-1}] +  {(1 + z + zy)}^{ - 1} }

→\;\bold{{(1 + z + zy)}^{ - 1} (yz+z+1)}

→\;{\bold{\frac{1}{(1+z+zy)}+(1+z+zy)}}

→\;{\bold{\cancel{\frac{(1+z+zy)}{(1+z+zy)}}}}

→\;\red{\bold{1}}

  • Hence Proved!!

\boxed{\tt{@MrSoverign}}

Hope This Helps!!

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